Indefinite IntegrationeasyFree

Integral of (cosx + x sinx)/(x²(cosx/x − 1)) | JEE

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

Question
∫cos⁡x+xsin⁡xx2(cos⁡xx−1) dx\displaystyle\int \dfrac{\cos x+x\sin x}{x^{2}\left(\dfrac{\cos x}{x}-1\right)}\,dx is equal to
Aln⁡∣Cxcos⁡x−x∣\ln\left|\dfrac{Cx}{\cos x-x}\right|correct
Bln⁡∣x∣+ln⁡∣cos⁡x+x∣+2ln⁡∣cos⁡x∣+C\ln|x|+\ln|\cos x+x|+2\ln|\cos x|+C
Cln⁡∣cos⁡x−xCx∣\ln\left|\dfrac{\cos x-x}{Cx}\right|
D−ln⁡∣x∣+ln⁡∣cos⁡x+x∣+C-\ln|x|+\ln|\cos x+x|+C
Solution
Step 1: Substitute t=cos⁡xx−1t=\dfrac{\cos x}{x}-1. Step 2: Differentiate, applying the quotient rule to cos⁡xx\dfrac{\cos x}{x}:
dt=(−sin⁡x) x−cos⁡xx2 dx=−xsin⁡x+cos⁡xx2 dxdt=\dfrac{(-\sin x)\,x-\cos x}{x^{2}}\,dx=-\dfrac{x\sin x+\cos x}{x^{2}}\,dx
⇒cos⁡x+xsin⁡xx2 dx=− dt\Rightarrow \dfrac{\cos x+x\sin x}{x^{2}}\,dx=-\,dt
 Step 3: The integral becomes
I=∫−dtt=−ln⁡∣t∣+KI=\int\dfrac{-dt}{t}=-\ln|t|+K
 Step 4: Back-substitute t=cos⁡xx−1=cos⁡x−xxt=\dfrac{\cos x}{x}-1=\dfrac{\cos x-x}{x}:
I=−ln⁡∣cos⁡x−xx∣+K=ln⁡∣xcos⁡x−x∣+K=ln⁡∣Cxcos⁡x−x∣I=-\ln\left|\dfrac{\cos x-x}{x}\right|+K=\ln\left|\dfrac{x}{\cos x-x}\right|+K=\ln\left|\dfrac{Cx}{\cos x-x}\right|
 Correct answer: (1)
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