Where Mindful Learning Begins
Where Mindful Learning Begins
Chapter 7 Ex 7.6 covers integration by parts — all 24 questions solved step by step, using the ILATE rule to pick the first function, the e^x[f(x)+f'(x)] shortcut, and a double integration-by-parts trick for products involving e^{2x}.
The hardest part of this exercise is rarely the integration itself — it's picking which function to differentiate first. ILATE (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) gives you that order of priority, and getting it backwards usually turns a two-line solution into a dead end. Questions 1–15 are direct applications building comfort with the rule, including the special case where the "second function" is just 1 (Q4, Q13). From Q16 onward, watch for the pattern e^x[f(x)+f'(x)], which integrates in one step to e^x f(x) without any further work — several questions are built specifically to test whether you can spot it instead of grinding through integration by parts twice. The final MCQs check both skills together.
Take u=x,\,dv=\sin x\,dx\Rightarrow du=dx,\,v=-\cos x. By parts: \displaystyle\int x\sin x\,dx = -x\cos x + \int \cos x\,dx.
Take u=x,\,dv=\sin3x\,dx\Rightarrow du=dx,\,v=-\dfrac{\cos3x}{3}. By parts: \displaystyle\int x\sin3x\,dx = -\dfrac{x\cos3x}{3} + \dfrac13\int \cos3x\,dx.
Take u=x^2,\,dv=e^x\,dx\Rightarrow du=2x\,dx,\,v=e^x: \displaystyle\int x^2e^x\,dx = x^2e^x - 2\int xe^x\,dx. Apply integration by parts once more to \int xe^x\,dx, giving xe^x-e^x.
Take u=\log x,\,dv=x\,dx\Rightarrow du=\dfrac{1}{x}dx,\,v=\dfrac{x^2}{2}. By parts: \displaystyle\int x\log x\,dx = \dfrac{x^2}{2}\log x - \int \dfrac{x}{2}\,dx.
Take u=\log2x,\,dv=x\,dx; since \dfrac{d}{dx}\log2x=\dfrac{1}{x} as well, the same computation as Q4 applies.
Take u=\log x,\,dv=x^2\,dx\Rightarrow du=\dfrac{1}{x}dx,\,v=\dfrac{x^3}{3}. By parts: \displaystyle\int x^2\log x\,dx = \dfrac{x^3}{3}\log x - \int \dfrac{x^2}{3}\,dx.
Take u=\sin^{-1}x,\,dv=x\,dx\Rightarrow du=\dfrac{1}{\sqrt{1-x^2}}dx,\,v=\dfrac{x^2}{2}: \displaystyle\int x\sin^{-1}x\,dx = \dfrac{x^2}{2}\sin^{-1}x - \dfrac12\int \dfrac{x^2}{\sqrt{1-x^2}}\,dx — call this integral (1).
Rewrite (1) using \dfrac{x^2}{\sqrt{1-x^2}} = \dfrac{1}{\sqrt{1-x^2}}-\sqrt{1-x^2}, which gives (1) = \dfrac12\sin^{-1}x - \dfrac{x}{2}\sqrt{1-x^2}.
Take u=\tan^{-1}x,\,dv=x\,dx\Rightarrow du=\dfrac{1}{1+x^2}dx,\,v=\dfrac{x^2}{2}: \displaystyle\int x\tan^{-1}x\,dx = \dfrac{x^2}{2}\tan^{-1}x - \dfrac12\int \dfrac{x^2}{1+x^2}\,dx.
Since \dfrac{x^2}{1+x^2}=1-\dfrac{1}{1+x^2}, the remaining integral is x-\tan^{-1}x.
Take u=\cos^{-1}x,\,dv=x\,dx\Rightarrow du=-\dfrac{1}{\sqrt{1-x^2}}dx,\,v=\dfrac{x^2}{2}: \displaystyle\int x\cos^{-1}x\,dx = \dfrac{x^2}{2}\cos^{-1}x + \dfrac12\int \dfrac{x^2}{\sqrt{1-x^2}}\,dx.
As in Q7, this integral equals \dfrac12\sin^{-1}x-\dfrac{x}{2}\sqrt{1-x^2}.
Take u=(\sin^{-1}x)^2,\,dv=dx\Rightarrow du=\dfrac{2\sin^{-1}x}{\sqrt{1-x^2}}dx,\,v=x: \displaystyle\int (\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 - 2\int \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx — call the remaining integral (1).
For (1), substitute t=\sin^{-1}x so x=\sin t: (1)=\displaystyle\int t\sin t\,dt = -t\cos t+\sin t = -\sin^{-1}x\sqrt{1-x^2}+x.
Take u=\cos^{-1}x and dv=\dfrac{x}{\sqrt{1-x^2}}dx, so v=-\sqrt{1-x^2} and du=-\dfrac{1}{\sqrt{1-x^2}}dx.
By parts: \displaystyle\int \dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}\,\cos^{-1}x - \int 1\,dx.
Take u=x,\,dv=\sec^2x\,dx\Rightarrow du=dx,\,v=\tan x. By parts: \displaystyle\int x\sec^2x\,dx = x\tan x - \int \tan x\,dx.
Take u=\tan^{-1}x,\,dv=dx\Rightarrow du=\dfrac{1}{1+x^2}dx,\,v=x. By parts: \displaystyle\int \tan^{-1}x\,dx = x\tan^{-1}x - \int \dfrac{x}{1+x^2}\,dx.
Take u=(\log x)^2,\,dv=x\,dx\Rightarrow du=\dfrac{2\log x}{x}dx,\,v=\dfrac{x^2}{2}: \displaystyle\int x(\log x)^2\,dx = \dfrac{x^2}{2}(\log x)^2 - \int x\log x\,dx.
From Q4, \displaystyle\int x\log x\,dx = \dfrac{x^2}{2}\log x - \dfrac{x^2}{4}.
Split into \displaystyle\int x^2\log x\,dx + \int \log x\,dx.
From Q6, the first is \dfrac{x^3}{3}\log x - \dfrac{x^3}{9}; the second is the standard result x\log x-x (via u=\log x,\,dv=dx).
This matches the shortcut form e^x[f(x)+f'(x)] with f(x)=\sin x, since f'(x)=\cos x, so \displaystyle\int e^x[f(x)+f'(x)]\,dx = e^xf(x)+C applies directly.
Write x=(1+x)-1, so \dfrac{x}{(1+x)^2} = \dfrac{1}{1+x} - \dfrac{1}{(1+x)^2} — this matches f(x)+f'(x) with f(x)=\dfrac{1}{1+x}.
Using half-angle identities, \dfrac{1+\sin x}{1+\cos x} = \dfrac12\sec^2\left(\dfrac{x}{2}\right)+\tan\left(\dfrac{x}{2}\right) — this matches f(x)+f'(x) with f(x)=\tan\left(\dfrac{x}{2}\right).
This matches f(x)+f'(x) directly with f(x)=\dfrac{1}{x}, since f'(x)=-\dfrac{1}{x^2}.
Write x-3=(x-1)-2, so \dfrac{x-3}{(x-1)^3} = \dfrac{1}{(x-1)^2} - \dfrac{2}{(x-1)^3} — matches f(x)+f'(x) with f(x)=\dfrac{1}{(x-1)^2}.
Let I=\displaystyle\int e^{2x}\sin x\,dx. Taking u=\sin x,\,dv=e^{2x}dx: I = \dfrac{e^{2x}}{2}\sin x - \dfrac12\int e^{2x}\cos x\,dx — call the remaining integral J.
Apply integration by parts again to J with u=\cos x,\,dv=e^{2x}dx: J = \dfrac{e^{2x}}{2}\cos x + \dfrac12 I.
Substituting back gives \dfrac54 I = \dfrac{e^{2x}}{2}\sin x - \dfrac{e^{2x}}{4}\cos x.
Substituting x=\tan\theta gives \dfrac{2x}{1+x^2}=\sin2\theta, so \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=2\tan^{-1}x over the relevant domain, reducing the integral to 2\displaystyle\int \tan^{-1}x\,dx.
From Q13, this is 2\left[x\tan^{-1}x-\dfrac12\log(1+x^2)\right].
Put t=x^3, so dt=3x^2\,dx: the integral becomes \dfrac13\displaystyle\int e^t\,dt = \dfrac13 e^t.
Expand: \sec x(1+\tan x) = \sec x+\sec x\tan x — matches f(x)+f'(x) with f(x)=\sec x.
Expert CBSE Coaching · Class 9–12