- Ex. 7.1
- Ex. 7.2
- Ex. 7.2
- Ex. 7.3
- Ex. 7.4
- Ex. 7.5
- Ex. 7.6
- Ex. 7.7
- Ex. 7.8
- Ex. 7.9
- Ex. 7.10
- Ex. 7.11
- Miscellaneous Exercise
Integrals Class 12th Mathematics Part II CBSE Solution - Exercise 7.2
This exercise focuses on the method of integration by substitution. Below are the step-by-step solutions for every question.
Question 1: Integrate the function \( \frac{2x}{1+x^2} \)
Let \( 1+x^2 = t \)
Differentiating both sides with respect to x:
\( 2x dx = dt \)
Now substituting in the integral:
\( \int \frac{2x}{1+x^2} dx = \int \frac{1}{t} dt \)
\( = \log|t| + C \)
Putting back \( t = 1+x^2 \):
\( = \log(1+x^2) + C \)
Differentiating both sides with respect to x:
\( 2x dx = dt \)
Now substituting in the integral:
\( \int \frac{2x}{1+x^2} dx = \int \frac{1}{t} dt \)
\( = \log|t| + C \)
Putting back \( t = 1+x^2 \):
\( = \log(1+x^2) + C \)
Question 2: Integrate the function \( \frac{(\log x)^2}{x} \)
Let \( \log x = t \)
Differentiating:
\( \frac{1}{x} dx = dt \)
Integral becomes:
\( \int t^2 dt = \frac{t^3}{3} + C \)
Substituting \( t = \log x \):
\( = \frac{(\log x)^3}{3} + C \)
Differentiating:
\( \frac{1}{x} dx = dt \)
Integral becomes:
\( \int t^2 dt = \frac{t^3}{3} + C \)
Substituting \( t = \log x \):
\( = \frac{(\log x)^3}{3} + C \)
Question 3: Integrate the function \( \frac{1}{x + x \log x} \)
Rewrite the expression: \( \frac{1}{x(1 + \log x)} \)
Let \( 1 + \log x = t \)
\( \Rightarrow \frac{1}{x} dx = dt \)
Integral becomes:
\( \int \frac{1}{t} dt = \log|t| + C \)
\( = \log|1 + \log x| + C \)
Let \( 1 + \log x = t \)
\( \Rightarrow \frac{1}{x} dx = dt \)
Integral becomes:
\( \int \frac{1}{t} dt = \log|t| + C \)
\( = \log|1 + \log x| + C \)
Question 4: Integrate the function \( \sin x \sin(\cos x) \)
Let \( \cos x = t \)
\( \Rightarrow -\sin x dx = dt \Rightarrow \sin x dx = -dt \)
Integral becomes:
\( \int \sin(t) (-dt) = -\int \sin t dt \)
\( = -(-\cos t) + C = \cos t + C \)
\( = \cos(\cos x) + C \)
\( \Rightarrow -\sin x dx = dt \Rightarrow \sin x dx = -dt \)
Integral becomes:
\( \int \sin(t) (-dt) = -\int \sin t dt \)
\( = -(-\cos t) + C = \cos t + C \)
\( = \cos(\cos x) + C \)
Question 5: Integrate the function \( \sin(ax + b) \cos(ax + b) \)
We know that \( \sin 2A = 2 \sin A \cos A \), so \( \sin A \cos A = \frac{1}{2} \sin 2A \).
\( \int \sin(ax+b)\cos(ax+b) dx = \frac{1}{2} \int \sin(2ax + 2b) dx \)
Let \( 2ax + 2b = t \Rightarrow 2a dx = dt \Rightarrow dx = \frac{dt}{2a} \)
\( = \frac{1}{2} \int \sin t \frac{dt}{2a} = \frac{1}{4a} (-\cos t) + C \)
\( = -\frac{1}{4a} \cos(2ax + 2b) + C \)
\( \int \sin(ax+b)\cos(ax+b) dx = \frac{1}{2} \int \sin(2ax + 2b) dx \)
Let \( 2ax + 2b = t \Rightarrow 2a dx = dt \Rightarrow dx = \frac{dt}{2a} \)
\( = \frac{1}{2} \int \sin t \frac{dt}{2a} = \frac{1}{4a} (-\cos t) + C \)
\( = -\frac{1}{4a} \cos(2ax + 2b) + C \)
Question 6: Integrate the function \( \sqrt{ax + b} \)
Let \( ax + b = t \)
\( \Rightarrow a dx = dt \Rightarrow dx = \frac{dt}{a} \)
\( \int t^{1/2} \frac{dt}{a} = \frac{1}{a} \left( \frac{t^{3/2}}{3/2} \right) + C \)
\( = \frac{2}{3a} (ax + b)^{3/2} + C \)
\( \Rightarrow a dx = dt \Rightarrow dx = \frac{dt}{a} \)
\( \int t^{1/2} \frac{dt}{a} = \frac{1}{a} \left( \frac{t^{3/2}}{3/2} \right) + C \)
\( = \frac{2}{3a} (ax + b)^{3/2} + C \)
Question 7: Integrate the function \( x\sqrt{x+2} \)
Let \( x + 2 = t \Rightarrow x = t - 2 \)
\( dx = dt \)
Integral becomes:
\( \int (t-2)\sqrt{t} dt = \int (t^{3/2} - 2t^{1/2}) dt \)
\( = \frac{t^{5/2}}{5/2} - 2\frac{t^{3/2}}{3/2} + C \)
\( = \frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2} + C \)
\( dx = dt \)
Integral becomes:
\( \int (t-2)\sqrt{t} dt = \int (t^{3/2} - 2t^{1/2}) dt \)
\( = \frac{t^{5/2}}{5/2} - 2\frac{t^{3/2}}{3/2} + C \)
\( = \frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2} + C \)
Question 8: Integrate the function \( x\sqrt{1+2x^2} \)
Let \( 1 + 2x^2 = t \)
\( \Rightarrow 4x dx = dt \Rightarrow x dx = \frac{dt}{4} \)
Integral becomes:
\( \int \sqrt{t} \frac{dt}{4} = \frac{1}{4} \int t^{1/2} dt \)
\( = \frac{1}{4} \cdot \frac{2}{3} t^{3/2} + C \)
\( = \frac{1}{6} (1+2x^2)^{3/2} + C \)
\( \Rightarrow 4x dx = dt \Rightarrow x dx = \frac{dt}{4} \)
Integral becomes:
\( \int \sqrt{t} \frac{dt}{4} = \frac{1}{4} \int t^{1/2} dt \)
\( = \frac{1}{4} \cdot \frac{2}{3} t^{3/2} + C \)
\( = \frac{1}{6} (1+2x^2)^{3/2} + C \)
Question 9: Integrate the function \( (4x+2)\sqrt{x^2+x+1} \)
Factor out 2: \( 2 \int (2x+1)\sqrt{x^2+x+1} dx \)
Let \( x^2+x+1 = t \)
\( \Rightarrow (2x+1) dx = dt \)
Integral becomes:
\( 2 \int \sqrt{t} dt = 2 \cdot \frac{2}{3} t^{3/2} + C \)
\( = \frac{4}{3} (x^2+x+1)^{3/2} + C \)
Let \( x^2+x+1 = t \)
\( \Rightarrow (2x+1) dx = dt \)
Integral becomes:
\( 2 \int \sqrt{t} dt = 2 \cdot \frac{2}{3} t^{3/2} + C \)
\( = \frac{4}{3} (x^2+x+1)^{3/2} + C \)
Question 10: Integrate the function \( \frac{1}{x - \sqrt{x}} \)
Rewrite: \( \frac{1}{\sqrt{x}(\sqrt{x} - 1)} \)
Let \( \sqrt{x} - 1 = t \)
\( \Rightarrow \frac{1}{2\sqrt{x}} dx = dt \Rightarrow \frac{1}{\sqrt{x}} dx = 2dt \)
Integral becomes:
\( \int \frac{2dt}{t} = 2 \log|t| + C \)
\( = 2 \log|\sqrt{x} - 1| + C \)
Let \( \sqrt{x} - 1 = t \)
\( \Rightarrow \frac{1}{2\sqrt{x}} dx = dt \Rightarrow \frac{1}{\sqrt{x}} dx = 2dt \)
Integral becomes:
\( \int \frac{2dt}{t} = 2 \log|t| + C \)
\( = 2 \log|\sqrt{x} - 1| + C \)
Question 11: Integrate the function \( \frac{x}{\sqrt{x+4}}, x > 0 \)
Let \( x + 4 = t \Rightarrow x = t - 4 \)
\( dx = dt \)
Integral becomes:
\( \int \frac{t-4}{\sqrt{t}} dt = \int (t^{1/2} - 4t^{-1/2}) dt \)
\( = \frac{2}{3} t^{3/2} - 4(2t^{1/2}) + C \)
\( = \frac{2}{3} (x+4)^{3/2} - 8\sqrt{x+4} + C \)
\( dx = dt \)
Integral becomes:
\( \int \frac{t-4}{\sqrt{t}} dt = \int (t^{1/2} - 4t^{-1/2}) dt \)
\( = \frac{2}{3} t^{3/2} - 4(2t^{1/2}) + C \)
\( = \frac{2}{3} (x+4)^{3/2} - 8\sqrt{x+4} + C \)
Question 12: Integrate the function \( (x^3 - 1)^{1/3}x^5 \)
Break \( x^5 \) into \( x^3 \cdot x^2 \): \( \int (x^3-1)^{1/3} x^3 \cdot x^2 dx \)
Let \( x^3 - 1 = t \Rightarrow x^3 = t+1 \)
\( \Rightarrow 3x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{3} \)
Integral becomes:
\( \int t^{1/3}(t+1) \frac{dt}{3} = \frac{1}{3} \int (t^{4/3} + t^{1/3}) dt \)
\( = \frac{1}{3} \left[ \frac{3}{7}t^{7/3} + \frac{3}{4}t^{4/3} \right] + C \)
\( = \frac{1}{7}(x^3-1)^{7/3} + \frac{1}{4}(x^3-1)^{4/3} + C \)
Let \( x^3 - 1 = t \Rightarrow x^3 = t+1 \)
\( \Rightarrow 3x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{3} \)
Integral becomes:
\( \int t^{1/3}(t+1) \frac{dt}{3} = \frac{1}{3} \int (t^{4/3} + t^{1/3}) dt \)
\( = \frac{1}{3} \left[ \frac{3}{7}t^{7/3} + \frac{3}{4}t^{4/3} \right] + C \)
\( = \frac{1}{7}(x^3-1)^{7/3} + \frac{1}{4}(x^3-1)^{4/3} + C \)
Question 13: Integrate the function \( \frac{x^2}{(2+3x^3)^3} \)
Let \( 2 + 3x^3 = t \)
\( \Rightarrow 9x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{9} \)
Integral becomes:
\( \int \frac{1}{t^3} \frac{dt}{9} = \frac{1}{9} \int t^{-3} dt \)
\( = \frac{1}{9} \left( \frac{t^{-2}}{-2} \right) + C \)
\( = -\frac{1}{18(2+3x^3)^2} + C \)
\( \Rightarrow 9x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{9} \)
Integral becomes:
\( \int \frac{1}{t^3} \frac{dt}{9} = \frac{1}{9} \int t^{-3} dt \)
\( = \frac{1}{9} \left( \frac{t^{-2}}{-2} \right) + C \)
\( = -\frac{1}{18(2+3x^3)^2} + C \)
Question 14: Integrate the function \( \frac{1}{x(\log x)^m}, x>0, m \neq 1 \)
Let \( \log x = t \)
\( \Rightarrow \frac{1}{x} dx = dt \)
Integral becomes:
\( \int \frac{dt}{t^m} = \int t^{-m} dt \)
\( = \frac{t^{1-m}}{1-m} + C \)
\( = \frac{(\log x)^{1-m}}{1-m} + C \)
\( \Rightarrow \frac{1}{x} dx = dt \)
Integral becomes:
\( \int \frac{dt}{t^m} = \int t^{-m} dt \)
\( = \frac{t^{1-m}}{1-m} + C \)
\( = \frac{(\log x)^{1-m}}{1-m} + C \)
Question 15: Integrate the function \( \frac{x}{9-4x^2} \)
Let \( 9 - 4x^2 = t \)
\( \Rightarrow -8x dx = dt \Rightarrow x dx = -\frac{dt}{8} \)
Integral becomes:
\( -\frac{1}{8} \int \frac{dt}{t} = -\frac{1}{8} \log|t| + C \)
\( = -\frac{1}{8} \log|9-4x^2| + C \)
\( \Rightarrow -8x dx = dt \Rightarrow x dx = -\frac{dt}{8} \)
Integral becomes:
\( -\frac{1}{8} \int \frac{dt}{t} = -\frac{1}{8} \log|t| + C \)
\( = -\frac{1}{8} \log|9-4x^2| + C \)
Question 16: Integrate the function \( e^{2x+3} \)
Let \( 2x + 3 = t \Rightarrow 2 dx = dt \)
\( \int e^t \frac{dt}{2} = \frac{1}{2} e^t + C \)
\( = \frac{1}{2} e^{2x+3} + C \)
\( \int e^t \frac{dt}{2} = \frac{1}{2} e^t + C \)
\( = \frac{1}{2} e^{2x+3} + C \)
Question 17: Integrate the function \( \frac{x}{e^{x^2}} \)
Let \( x^2 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2} \)
Integral becomes:
\( \int \frac{1}{e^t} \frac{dt}{2} = \frac{1}{2} \int e^{-t} dt \)
\( = \frac{1}{2} (-e^{-t}) + C \)
\( = -\frac{1}{2} e^{-x^2} + C \)
Integral becomes:
\( \int \frac{1}{e^t} \frac{dt}{2} = \frac{1}{2} \int e^{-t} dt \)
\( = \frac{1}{2} (-e^{-t}) + C \)
\( = -\frac{1}{2} e^{-x^2} + C \)
Question 18: Integrate the function \( \frac{e^{\tan^{-1}x}}{1+x^2} \)
Let \( \tan^{-1}x = t \)
\( \Rightarrow \frac{1}{1+x^2} dx = dt \)
Integral becomes:
\( \int e^t dt = e^t + C \)
\( = e^{\tan^{-1}x} + C \)
\( \Rightarrow \frac{1}{1+x^2} dx = dt \)
Integral becomes:
\( \int e^t dt = e^t + C \)
\( = e^{\tan^{-1}x} + C \)
Question 19: Integrate the function \( \frac{e^{2x}-1}{e^{2x}+1} \)
Divide numerator and denominator by \( e^x \):
\( \frac{e^x - e^{-x}}{e^x + e^{-x}} \)
Let \( e^x + e^{-x} = t \)
\( \Rightarrow (e^x - e^{-x}) dx = dt \)
Integral becomes:
\( \int \frac{dt}{t} = \log|t| + C \)
\( = \log(e^x + e^{-x}) + C \)
\( \frac{e^x - e^{-x}}{e^x + e^{-x}} \)
Let \( e^x + e^{-x} = t \)
\( \Rightarrow (e^x - e^{-x}) dx = dt \)
Integral becomes:
\( \int \frac{dt}{t} = \log|t| + C \)
\( = \log(e^x + e^{-x}) + C \)
Question 20: Integrate the function \( \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \)
Let \( e^{2x} + e^{-2x} = t \)
\( \Rightarrow (2e^{2x} - 2e^{-2x}) dx = dt \)
\( \Rightarrow 2(e^{2x} - e^{-2x}) dx = dt \Rightarrow (e^{2x} - e^{-2x}) dx = \frac{dt}{2} \)
Integral becomes:
\( \int \frac{1}{t} \frac{dt}{2} = \frac{1}{2} \log|t| + C \)
\( = \frac{1}{2} \log(e^{2x} + e^{-2x}) + C \)
\( \Rightarrow (2e^{2x} - 2e^{-2x}) dx = dt \)
\( \Rightarrow 2(e^{2x} - e^{-2x}) dx = dt \Rightarrow (e^{2x} - e^{-2x}) dx = \frac{dt}{2} \)
Integral becomes:
\( \int \frac{1}{t} \frac{dt}{2} = \frac{1}{2} \log|t| + C \)
\( = \frac{1}{2} \log(e^{2x} + e^{-2x}) + C \)
Question 21: Integrate the function \( \tan^2(2x-3) \)
Use identity: \( \tan^2 \theta = \sec^2 \theta - 1 \)
\( \int (\sec^2(2x-3) - 1) dx \)
\( = \int \sec^2(2x-3) dx - \int 1 dx \)
For the first part, let \( 2x-3 = t \Rightarrow 2dx = dt \)
\( \frac{1}{2} \tan(2x-3) - x + C \)
\( \int (\sec^2(2x-3) - 1) dx \)
\( = \int \sec^2(2x-3) dx - \int 1 dx \)
For the first part, let \( 2x-3 = t \Rightarrow 2dx = dt \)
\( \frac{1}{2} \tan(2x-3) - x + C \)
Question 22: Integrate the function \( \sec^2(7-4x) \)
Let \( 7 - 4x = t \Rightarrow -4 dx = dt \Rightarrow dx = -\frac{dt}{4} \)
\( \int \sec^2 t \frac{-dt}{4} = -\frac{1}{4} \tan t + C \)
\( = -\frac{1}{4} \tan(7-4x) + C \)
\( \int \sec^2 t \frac{-dt}{4} = -\frac{1}{4} \tan t + C \)
\( = -\frac{1}{4} \tan(7-4x) + C \)
Question 23: Integrate the function \( \frac{\sin^{-1}x}{\sqrt{1-x^2}} \)
Let \( \sin^{-1}x = t \)
\( \Rightarrow \frac{1}{\sqrt{1-x^2}} dx = dt \)
Integral becomes:
\( \int t dt = \frac{t^2}{2} + C \)
\( = \frac{(\sin^{-1}x)^2}{2} + C \)
\( \Rightarrow \frac{1}{\sqrt{1-x^2}} dx = dt \)
Integral becomes:
\( \int t dt = \frac{t^2}{2} + C \)
\( = \frac{(\sin^{-1}x)^2}{2} + C \)
Question 24: Integrate the function \( \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} \)
Factor denominator: \( 2(3\cos x + 2\sin x) \)
Let \( 3\cos x + 2\sin x = t \)
\( \Rightarrow (-3\sin x + 2\cos x) dx = dt \)
The numerator is exactly \( dt \).
Integral becomes:
\( \int \frac{dt}{2t} = \frac{1}{2} \log|t| + C \)
\( = \frac{1}{2} \log|3\cos x + 2\sin x| + C \)
Let \( 3\cos x + 2\sin x = t \)
\( \Rightarrow (-3\sin x + 2\cos x) dx = dt \)
The numerator is exactly \( dt \).
Integral becomes:
\( \int \frac{dt}{2t} = \frac{1}{2} \log|t| + C \)
\( = \frac{1}{2} \log|3\cos x + 2\sin x| + C \)
Question 25: Integrate the function \( \frac{1}{\cos^2x (1-\tan x)^2} \)
Rewrite as \( \frac{\sec^2 x}{(1-\tan x)^2} \)
Let \( 1 - \tan x = t \)
\( \Rightarrow -\sec^2 x dx = dt \Rightarrow \sec^2 x dx = -dt \)
Integral becomes:
\( \int \frac{-dt}{t^2} = -\int t^{-2} dt \)
\( = -(\frac{t^{-1}}{-1}) + C = \frac{1}{t} + C \)
\( = \frac{1}{1-\tan x} + C \)
Let \( 1 - \tan x = t \)
\( \Rightarrow -\sec^2 x dx = dt \Rightarrow \sec^2 x dx = -dt \)
Integral becomes:
\( \int \frac{-dt}{t^2} = -\int t^{-2} dt \)
\( = -(\frac{t^{-1}}{-1}) + C = \frac{1}{t} + C \)
\( = \frac{1}{1-\tan x} + C \)
Question 26: Integrate the function \( \frac{\cos\sqrt{x}}{\sqrt{x}} \)
Let \( \sqrt{x} = t \)
\( \Rightarrow \frac{1}{2\sqrt{x}} dx = dt \Rightarrow \frac{1}{\sqrt{x}} dx = 2dt \)
Integral becomes:
\( \int \cos t (2dt) = 2 \sin t + C \)
\( = 2 \sin\sqrt{x} + C \)
\( \Rightarrow \frac{1}{2\sqrt{x}} dx = dt \Rightarrow \frac{1}{\sqrt{x}} dx = 2dt \)
Integral becomes:
\( \int \cos t (2dt) = 2 \sin t + C \)
\( = 2 \sin\sqrt{x} + C \)
Question 27: Integrate the function \( \sqrt{\sin 2x} \cos 2x \)
Let \( \sin 2x = t \)
\( \Rightarrow 2 \cos 2x dx = dt \Rightarrow \cos 2x dx = \frac{dt}{2} \)
Integral becomes:
\( \int \sqrt{t} \frac{dt}{2} = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + C \)
\( = \frac{1}{3} (\sin 2x)^{3/2} + C \)
\( \Rightarrow 2 \cos 2x dx = dt \Rightarrow \cos 2x dx = \frac{dt}{2} \)
Integral becomes:
\( \int \sqrt{t} \frac{dt}{2} = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + C \)
\( = \frac{1}{3} (\sin 2x)^{3/2} + C \)
Question 28: Integrate the function \( \frac{\cos x}{\sqrt{1+\sin x}} \)
Let \( 1 + \sin x = t \)
\( \Rightarrow \cos x dx = dt \)
Integral becomes:
\( \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt \)
\( = 2t^{1/2} + C = 2\sqrt{1+\sin x} + C \)
\( \Rightarrow \cos x dx = dt \)
Integral becomes:
\( \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt \)
\( = 2t^{1/2} + C = 2\sqrt{1+\sin x} + C \)
Question 29: Integrate the function \( \cot x \log(\sin x) \)
Let \( \log(\sin x) = t \)
\( \Rightarrow \frac{1}{\sin x} \cdot \cos x dx = dt \Rightarrow \cot x dx = dt \)
Integral becomes:
\( \int t dt = \frac{t^2}{2} + C \)
\( = \frac{1}{2} (\log(\sin x))^2 + C \)
\( \Rightarrow \frac{1}{\sin x} \cdot \cos x dx = dt \Rightarrow \cot x dx = dt \)
Integral becomes:
\( \int t dt = \frac{t^2}{2} + C \)
\( = \frac{1}{2} (\log(\sin x))^2 + C \)
Question 30: Integrate the function \( \frac{\sin x}{1+\cos x} \)
Let \( 1 + \cos x = t \)
\( \Rightarrow -\sin x dx = dt \Rightarrow \sin x dx = -dt \)
Integral becomes:
\( \int \frac{-dt}{t} = -\log|t| + C \)
\( = -\log|1+\cos x| + C \)
\( \Rightarrow -\sin x dx = dt \Rightarrow \sin x dx = -dt \)
Integral becomes:
\( \int \frac{-dt}{t} = -\log|t| + C \)
\( = -\log|1+\cos x| + C \)
Question 31: Integrate the function \( \frac{\sin x}{(1+\cos x)^2} \)
Let \( 1 + \cos x = t \)
\( \Rightarrow -\sin x dx = dt \)
Integral becomes:
\( \int \frac{-dt}{t^2} = -\int t^{-2} dt \)
\( = -(\frac{t^{-1}}{-1}) + C = \frac{1}{t} + C \)
\( = \frac{1}{1+\cos x} + C \)
\( \Rightarrow -\sin x dx = dt \)
Integral becomes:
\( \int \frac{-dt}{t^2} = -\int t^{-2} dt \)
\( = -(\frac{t^{-1}}{-1}) + C = \frac{1}{t} + C \)
\( = \frac{1}{1+\cos x} + C \)
Question 32: Integrate the function \( \frac{1}{1+\cot x} \)
\( I = \int \frac{1}{1+\frac{\cos x}{\sin x}} dx = \int \frac{\sin x}{\sin x + \cos x} dx \)
Multiply and divide by 2:
\( I = \frac{1}{2} \int \frac{2\sin x}{\sin x + \cos x} dx \)
\( = \frac{1}{2} \int \frac{(\sin x + \cos x) - (\cos x - \sin x)}{\sin x + \cos x} dx \)
\( = \frac{1}{2} \left[ \int 1 dx - \int \frac{\cos x - \sin x}{\sin x + \cos x} dx \right] \)
Let \( \sin x + \cos x = t \Rightarrow (\cos x - \sin x)dx = dt \) in second part.
\( = \frac{1}{2} [x - \log|\sin x + \cos x|] + C \)
Multiply and divide by 2:
\( I = \frac{1}{2} \int \frac{2\sin x}{\sin x + \cos x} dx \)
\( = \frac{1}{2} \int \frac{(\sin x + \cos x) - (\cos x - \sin x)}{\sin x + \cos x} dx \)
\( = \frac{1}{2} \left[ \int 1 dx - \int \frac{\cos x - \sin x}{\sin x + \cos x} dx \right] \)
Let \( \sin x + \cos x = t \Rightarrow (\cos x - \sin x)dx = dt \) in second part.
\( = \frac{1}{2} [x - \log|\sin x + \cos x|] + C \)
Question 33: Integrate the function \( \frac{1}{1-\tan x} \)
\( I = \int \frac{1}{1-\frac{\sin x}{\cos x}} dx = \int \frac{\cos x}{\cos x - \sin x} dx \)
Multiply and divide by 2:
\( I = \frac{1}{2} \int \frac{2\cos x}{\cos x - \sin x} dx \)
\( = \frac{1}{2} \int \frac{(\cos x - \sin x) + (\cos x + \sin x)}{\cos x - \sin x} dx \)
\( = \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\cos x + \sin x}{\cos x - \sin x} dx \)
In second integral, let \( \cos x - \sin x = t \Rightarrow (-\sin x - \cos x)dx = dt \Rightarrow (\sin x + \cos x)dx = -dt \)
\( = \frac{1}{2}x - \frac{1}{2}\log|\cos x - \sin x| + C \)
Multiply and divide by 2:
\( I = \frac{1}{2} \int \frac{2\cos x}{\cos x - \sin x} dx \)
\( = \frac{1}{2} \int \frac{(\cos x - \sin x) + (\cos x + \sin x)}{\cos x - \sin x} dx \)
\( = \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\cos x + \sin x}{\cos x - \sin x} dx \)
In second integral, let \( \cos x - \sin x = t \Rightarrow (-\sin x - \cos x)dx = dt \Rightarrow (\sin x + \cos x)dx = -dt \)
\( = \frac{1}{2}x - \frac{1}{2}\log|\cos x - \sin x| + C \)
Question 34: Integrate the function \( \frac{\sqrt{\tan x}}{\sin x \cos x} \)
Multiply numerator and denominator by \( \cos x \):
\( = \int \frac{\sqrt{\tan x} \cos x}{\sin x \cos^2 x} dx = \int \frac{\sqrt{\tan x}}{\tan x \cos^2 x} dx \)
\( = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx \)
Let \( \tan x = t \Rightarrow \sec^2 x dx = dt \)
\( = \int t^{-1/2} dt = 2\sqrt{t} + C \)
\( = 2\sqrt{\tan x} + C \)
\( = \int \frac{\sqrt{\tan x} \cos x}{\sin x \cos^2 x} dx = \int \frac{\sqrt{\tan x}}{\tan x \cos^2 x} dx \)
\( = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx \)
Let \( \tan x = t \Rightarrow \sec^2 x dx = dt \)
\( = \int t^{-1/2} dt = 2\sqrt{t} + C \)
\( = 2\sqrt{\tan x} + C \)
Question 35: Integrate the function \( \frac{(1+\log x)^2}{x} \)
Let \( 1 + \log x = t \)
\( \Rightarrow \frac{1}{x} dx = dt \)
Integral becomes:
\( \int t^2 dt = \frac{t^3}{3} + C \)
\( = \frac{(1+\log x)^3}{3} + C \)
\( \Rightarrow \frac{1}{x} dx = dt \)
Integral becomes:
\( \int t^2 dt = \frac{t^3}{3} + C \)
\( = \frac{(1+\log x)^3}{3} + C \)
Question 36: Integrate the function \( \frac{(x+1)(x+\log x)^2}{x} \)
Rewrite as \( \left( \frac{x+1}{x} \right) (x+\log x)^2 = (1 + \frac{1}{x})(x+\log x)^2 \)
Let \( x + \log x = t \)
\( \Rightarrow (1 + \frac{1}{x}) dx = dt \)
Integral becomes:
\( \int t^2 dt = \frac{t^3}{3} + C \)
\( = \frac{(x+\log x)^3}{3} + C \)
Let \( x + \log x = t \)
\( \Rightarrow (1 + \frac{1}{x}) dx = dt \)
Integral becomes:
\( \int t^2 dt = \frac{t^3}{3} + C \)
\( = \frac{(x+\log x)^3}{3} + C \)
Question 37: Integrate the function \( \frac{x^3 \sin(\tan^{-1}x^4)}{1+x^8} \)
Let \( x^4 = t \Rightarrow 4x^3 dx = dt \)
Integral becomes \( \frac{1}{4} \int \frac{\sin(\tan^{-1}t)}{1+t^2} dt \)
Now let \( \tan^{-1}t = u \Rightarrow \frac{1}{1+t^2} dt = du \)
\( = \frac{1}{4} \int \sin u du = -\frac{1}{4} \cos u + C \)
\( = -\frac{1}{4} \cos(\tan^{-1}t) + C \)
\( = -\frac{1}{4} \cos(\tan^{-1}x^4) + C \)
Integral becomes \( \frac{1}{4} \int \frac{\sin(\tan^{-1}t)}{1+t^2} dt \)
Now let \( \tan^{-1}t = u \Rightarrow \frac{1}{1+t^2} dt = du \)
\( = \frac{1}{4} \int \sin u du = -\frac{1}{4} \cos u + C \)
\( = -\frac{1}{4} \cos(\tan^{-1}t) + C \)
\( = -\frac{1}{4} \cos(\tan^{-1}x^4) + C \)
Question 38: \( \int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} dx \) equals
Let \( x^{10} + 10^x = t \)
Differentiating: \( (10x^9 + 10^x \log_e 10) dx = dt \)
Integral becomes:
\( \int \frac{dt}{t} = \log|t| + C \)
\( = \log(x^{10} + 10^x) + C \)
Differentiating: \( (10x^9 + 10^x \log_e 10) dx = dt \)
Integral becomes:
\( \int \frac{dt}{t} = \log|t| + C \)
\( = \log(x^{10} + 10^x) + C \)
Question 39: \( \int \frac{dx}{\sin^2 x \cos^2 x} \) equals
Use identity \( 1 = \sin^2 x + \cos^2 x \) in numerator:
\( \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx \)
\( = \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} dx \)
\( = \int \sec^2 x dx + \int \text{cosec}^2 x dx \)
\( = \tan x - \cot x + C \)
\( \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx \)
\( = \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} dx \)
\( = \int \sec^2 x dx + \int \text{cosec}^2 x dx \)
\( = \tan x - \cot x + C \)