Download Exercise 7.6 PDF
- 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 7.6
This exercise deals with Integration by Parts. The formula used is: $$ \int f(x)g(x)dx = f(x)\int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx $$ The choice of the first function \(f(x)\) and second function \(g(x)\) is usually based on the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential).
Let \( I = \int x \sin x \, dx \).
Using integration by parts, let \( u = x \) (Algebraic) and \( v = \sin x \) (Trigonometric).
$$ I = x \int \sin x \, dx - \int \left( \frac{d}{dx}(x) \int \sin x \, dx \right) dx $$ $$ I = x(-\cos x) - \int 1 \cdot (-\cos x) \, dx $$ $$ I = -x \cos x + \int \cos x \, dx $$ $$ I = -x \cos x + \sin x + C $$Let \( I = \int x \sin 3x \, dx \).
Using integration by parts, take \( x \) as the first function and \( \sin 3x \) as the second function.
$$ I = x \int \sin 3x \, dx - \int \left( \frac{d}{dx}(x) \int \sin 3x \, dx \right) dx $$ $$ I = x\left(\frac{-\cos 3x}{3}\right) - \int 1 \cdot \left(\frac{-\cos 3x}{3}\right) dx $$ $$ I = -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx $$ $$ I = -\frac{x \cos 3x}{3} + \frac{1}{9} \sin 3x + C $$Let \( I = \int x^2 e^x \, dx \).
Applying integration by parts with \( u = x^2 \) and \( v = e^x \):
$$ I = x^2 \int e^x \, dx - \int \left( \frac{d}{dx}(x^2) \int e^x \, dx \right) dx $$ $$ I = x^2 e^x - \int 2x e^x \, dx $$ $$ I = x^2 e^x - 2 \int x e^x \, dx $$Applying integration by parts again for \( \int x e^x \, dx \):
$$ I = x^2 e^x - 2 \left[ x \int e^x \, dx - \int \left( \frac{d}{dx}(x) \int e^x \, dx \right) dx \right] $$ $$ I = x^2 e^x - 2 \left[ x e^x - \int e^x \, dx \right] $$ $$ I = x^2 e^x - 2(x e^x - e^x) + C $$ $$ I = e^x (x^2 - 2x + 2) + C $$Let \( I = \int x \log x \, dx \).
Using ILATE, take Logarithmic (\( \log x \)) as first function and Algebraic (\( x \)) as second function.
$$ I = \log x \int x \, dx - \int \left( \frac{d}{dx}(\log x) \int x \, dx \right) dx $$ $$ I = \log x \left(\frac{x^2}{2}\right) - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \log x}{2} - \frac{1}{2} \int x \, dx $$ $$ I = \frac{x^2 \log x}{2} - \frac{x^2}{4} + C $$Let \( I = \int x \log 2x \, dx \).
$$ I = \log 2x \int x \, dx - \int \left( \frac{d}{dx}(\log 2x) \int x \, dx \right) dx $$Note: \( \frac{d}{dx}(\log 2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \).
$$ I = \frac{x^2}{2} \log 2x - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \log 2x}{2} - \frac{1}{2} \int x \, dx $$ $$ I = \frac{x^2 \log 2x}{2} - \frac{x^2}{4} + C $$Let \( I = \int x^2 \log x \, dx \).
$$ I = \log x \int x^2 \, dx - \int \left( \frac{d}{dx}(\log x) \int x^2 \, dx \right) dx $$ $$ I = \log x \left(\frac{x^3}{3}\right) - \int \frac{1}{x} \cdot \frac{x^3}{3} \, dx $$ $$ I = \frac{x^3 \log x}{3} - \frac{1}{3} \int x^2 \, dx $$ $$ I = \frac{x^3 \log x}{3} - \frac{x^3}{9} + C $$Let \( I = \int x \sin^{-1} x \, dx \).
Take \( \sin^{-1} x \) as first function.
$$ I = \sin^{-1} x \left(\frac{x^2}{2}\right) - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \left( \sqrt{1-x^2} - \frac{1}{\sqrt{1-x^2}} \right) dx $$Using standard integrals \( \int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \).
$$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x - \sin^{-1} x \right] + C $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{x}{4}\sqrt{1-x^2} - \frac{1}{4}\sin^{-1} x + C $$ $$ I = \frac{1}{4}(2x^2-1)\sin^{-1} x + \frac{x}{4}\sqrt{1-x^2} + C $$Let \( I = \int x \tan^{-1} x \, dx \).
$$ I = \tan^{-1} x \left(\frac{x^2}{2}\right) - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2+1-1}{1+x^2} \, dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \left( x - \tan^{-1} x \right) + C $$Let \( I = \int x \cos^{-1} x \, dx \).
Similar to Q7, but with \( \cos^{-1} x \).
$$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{-\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \left[ \int \sqrt{1-x^2} \, dx - \int \frac{1}{\sqrt{1-x^2}} \, dx \right] $$ $$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x - \sin^{-1} x \right] + C $$Since \( \sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x \), we can absorb constants.
Final simplified form:
$$ I = \frac{2x^2-1}{4}\cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + C $$Let \( I = \int (\sin^{-1} x)^2 \cdot 1 \, dx \).
$$ I = (\sin^{-1} x)^2 (x) - \int 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \cdot x \, dx $$For the integral part, put \( \sin^{-1} x = t \Rightarrow \frac{dx}{\sqrt{1-x^2}} = dt \) and \( x = \sin t \).
$$ \int 2t \sin t \, dt = 2 \left( t(-\cos t) - \int 1(-\cos t)dt \right) = -2t \cos t + 2 \sin t $$Substitute back \( t = \sin^{-1} x \), \( \cos t = \sqrt{1-x^2} \).
$$ I = x(\sin^{-1} x)^2 - [-2\sqrt{1-x^2}\sin^{-1} x + 2x] + C $$ $$ I = x(\sin^{-1} x)^2 + 2\sqrt{1-x^2}\sin^{-1} x - 2x + C $$Let \( I = \int \cos^{-1} x \cdot \frac{x}{\sqrt{1-x^2}} \, dx \).
Put \( \cos^{-1} x = t \Rightarrow \frac{-1}{\sqrt{1-x^2}} dx = dt \Rightarrow \frac{dx}{\sqrt{1-x^2}} = -dt \). Also \( x = \cos t \).
$$ I = \int t \cdot \cos t \cdot (-dt) = -\int t \cos t \, dt $$Integrating by parts:
$$ I = - [ t \sin t - \int \sin t \, dt ] $$ $$ I = - [ t \sin t + \cos t ] + C $$Substituting back \( t = \cos^{-1} x \), \( \sin t = \sqrt{1-x^2} \):
$$ I = - [\sqrt{1-x^2} \cos^{-1} x + x] + C $$Let \( I = \int x \sec^2 x \, dx \).
$$ I = x \int \sec^2 x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^2 x \, dx \right) dx $$ $$ I = x \tan x - \int \tan x \, dx $$ $$ I = x \tan x - \log|\sec x| + C $$Or \( I = x \tan x + \log|\cos x| + C \)
Let \( I = \int 1 \cdot \tan^{-1} x \, dx \).
$$ I = \tan^{-1} x (x) - \int \frac{1}{1+x^2} \cdot x \, dx $$ $$ I = x \tan^{-1} x - \frac{1}{2} \int \frac{2x}{1+x^2} \, dx $$ $$ I = x \tan^{-1} x - \frac{1}{2} \log|1+x^2| + C $$Let \( I = \int (\log x)^2 \cdot x \, dx \).
$$ I = (\log x)^2 \frac{x^2}{2} - \int 2 \log x \cdot \frac{1}{x} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2}{2} (\log x)^2 - \int x \log x \, dx $$Using result from Q4 for \( \int x \log x \, dx \):
$$ I = \frac{x^2}{2} (\log x)^2 - \left[ \frac{x^2}{2} \log x - \frac{x^2}{4} \right] + C $$ $$ I = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C $$Let \( I = \int \log x (x^2+1) \, dx \).
$$ I = \log x \left( \frac{x^3}{3} + x \right) - \int \frac{1}{x} \left( \frac{x^3}{3} + x \right) dx $$ $$ I = \left( \frac{x^3}{3} + x \right) \log x - \int \left( \frac{x^2}{3} + 1 \right) dx $$ $$ I = \left( \frac{x^3}{3} + x \right) \log x - \frac{x^3}{9} - x + C $$We use the property: \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).
Here, let \( f(x) = \sin x \), then \( f'(x) = \cos x \).
Thus, \( I = e^x \sin x + C \).
Rewrite integrand:
$$ I = \int e^x \left[ \frac{1+x-1}{(1+x)^2} \right] dx $$ $$ I = \int e^x \left[ \frac{1}{1+x} - \frac{1}{(1+x)^2} \right] dx $$Let \( f(x) = \frac{1}{1+x} \), then \( f'(x) = \frac{-1}{(1+x)^2} \).
Using the property, \( I = \frac{e^x}{1+x} + C \).
Using half-angle formulas:
$$ \frac{1+\sin x}{1+\cos x} = \frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} $$ $$ = \frac{1}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} $$ $$ = \frac{1}{2}\sec^2(x/2) + \tan(x/2) $$Integral becomes \( \int e^x [ \tan(x/2) + \frac{1}{2}\sec^2(x/2) ] dx \).
Let \( f(x) = \tan(x/2) \), then \( f'(x) = \frac{1}{2}\sec^2(x/2) \).
Thus, \( I = e^x \tan(x/2) + C \).
Let \( f(x) = \frac{1}{x} \), then \( f'(x) = -\frac{1}{x^2} \).
Using the standard property, \( I = \frac{e^x}{x} + C \).
Rewrite numerator \( x-3 \) as \( (x-1) - 2 \).
$$ I = \int e^x \left[ \frac{(x-1)-2}{(x-1)^3} \right] dx $$ $$ I = \int e^x \left[ \frac{1}{(x-1)^2} + \frac{-2}{(x-1)^3} \right] dx $$Let \( f(x) = (x-1)^{-2} \), then \( f'(x) = -2(x-1)^{-3} \).
Thus, \( I = \frac{e^x}{(x-1)^2} + C \).
Let \( I = \int e^{2x} \sin x \, dx \).
By parts: \( u=\sin x, v=e^{2x} \).
$$ I = \sin x \frac{e^{2x}}{2} - \int \cos x \frac{e^{2x}}{2} \, dx $$Apply parts again on integral:
$$ \int e^{2x} \cos x \, dx = \cos x \frac{e^{2x}}{2} - \int (-\sin x) \frac{e^{2x}}{2} \, dx $$Substitute back:
$$ I = \frac{e^{2x}\sin x}{2} - \frac{1}{2} \left[ \frac{e^{2x}\cos x}{2} + \frac{1}{2} I \right] $$ $$ I = \frac{e^{2x}\sin x}{2} - \frac{e^{2x}\cos x}{4} - \frac{1}{4} I $$ $$ \frac{5}{4} I = \frac{e^{2x}}{4} (2\sin x - \cos x) $$ $$ I = \frac{e^{2x}}{5} (2\sin x - \cos x) + C $$Put \( x = \tan \theta \implies dx = \sec^2 \theta \, d\theta \).
$$ \sin^{-1}(\sin 2\theta) = 2\theta $$ $$ I = \int 2\theta \sec^2 \theta \, d\theta $$By parts: \( 2 [ \theta \tan \theta - \int \tan \theta \, d\theta ] \).
$$ I = 2\theta \tan \theta - 2 \log|\sec \theta| + C $$Substitute \( \theta = \tan^{-1} x \). Note \( \log|\sec \theta| = \frac{1}{2}\log(1+\tan^2 \theta) = \frac{1}{2}\log(1+x^2) \).
$$ I = 2x \tan^{-1} x - \log(1+x^2) + C $$Let \( x^3 = t \implies 3x^2 dx = dt \implies x^2 dx = \frac{dt}{3} \).
$$ I = \int e^t \frac{dt}{3} = \frac{1}{3} e^t + C = \frac{1}{3} e^{x^3} + C $$Let \( f(x) = \sec x \), then \( f'(x) = \sec x \tan x \).
By standard property, \( I = e^x \sec x + C \).