Definite IntegrationmediumFree

Link ∫cosx/(x+2)² and ∫sin2x/(x+1) | JEE

JEE Maths question with a full step-by-step solution.

Question
If A=∫0πcos⁡x(x+2)2 dxA=\displaystyle\int_{0}^{\pi}\dfrac{\cos x}{(x+2)^{2}}\,dx, then ∫0π/2sin⁡2xx+1 dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin 2x}{x+1}\,dx is equal to
AA−12−1π+2A-\dfrac12-\dfrac{1}{\pi+2}
B12+1π+2−A\dfrac12+\dfrac{1}{\pi+2}-Acorrect
C1π+2−A\dfrac{1}{\pi+2}-A
D1+1π+2−A1+\dfrac{1}{\pi+2}-A
Solution
Step 1: Integrate AA by parts, taking u=cos⁡xu=\cos x and dv=dx(x+2)2dv=\dfrac{dx}{(x+2)^{2}} so that v=−1x+2v=-\dfrac{1}{x+2}:
A=[−cos⁡xx+2]0π−∫0πsin⁡xx+2 dx.A=\left[-\dfrac{\cos x}{x+2}\right]_{0}^{\pi}-\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx.
 The boundary term is −cos⁡ππ+2+cos⁡02=1π+2+12-\dfrac{\cos\pi}{\pi+2}+\dfrac{\cos 0}{2}=\dfrac{1}{\pi+2}+\dfrac12, hence
A=1π+2+12−∫0πsin⁡xx+2 dx.A=\dfrac{1}{\pi+2}+\dfrac12-\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx.
 Step 2: Transform the remaining integral. In ∫0πsin⁡xx+2 dx\displaystyle\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx put x=2tx=2t, dx=2 dtdx=2\,dt, with t:0→π2t:0\to\dfrac{\pi}{2}:
∫0πsin⁡xx+2 dx=∫0π/2sin⁡2t2t+2⋅2 dt=∫0π/2sin⁡2tt+1 dt.\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx=\int_{0}^{\pi/2}\dfrac{\sin 2t}{2t+2}\cdot 2\,dt=\int_{0}^{\pi/2}\dfrac{\sin 2t}{t+1}\,dt.
 Step 3: Substitute back into the expression for AA:
A=1π+2+12−∫0π/2sin⁡2tt+1 dt.A=\dfrac{1}{\pi+2}+\dfrac12-\int_{0}^{\pi/2}\dfrac{\sin 2t}{t+1}\,dt.
 Step 4: Solve for the required integral:
∫0π/2sin⁡2xx+1 dx=12+1π+2−A.\int_{0}^{\pi/2}\dfrac{\sin 2x}{x+1}\,dx=\dfrac12+\dfrac{1}{\pi+2}-A.
 Correct answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Definite Integration · easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Definite Integration · easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
Definite Integration · easy
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx, n∈Nn\in\mathbb{N}, then n=n=
Definite Integration · easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath