Indefinite IntegrationmediumFree

Integral of e^(sinx)(x·cscx − sec³x)/(cscx·secx) | JEE

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

Question

\displaystyle\int e^{\sin x}\,\dfrac{x\csc x-\sec^{3}x}{\csc x\,\sec x}\,dx

equals

e^{\sin x}-e^{\sin x}\sec x+C

x\,e^{\sin x}+e^{\sin x}\sec x+C

x\,e^{\sin x}-e^{\sin x}\sec x+C

correct

x\,e^{\sin x}-\sec x+C

Solution

Step 1: Simplify the integrand by dividing through by

\csc x\,\sec x

(i.e. multiplying by

\sin x\cos x

x\csc x\cdot\sin x\cos x=x\cos x,\qquad \sec^{3}x\cdot\sin x\cos x=\dfrac{\sin x}{\cos^{2}x}=\tan x\sec x

\Rightarrow \text{integrand}=e^{\sin x}\big(x\cos x-\tan x\sec x\big)

Step 2: Split into two integrals:

I=\int x\cos x\,e^{\sin x}\,dx-\int \tan x\sec x\,e^{\sin x}\,dx=I_{1}-I_{2}

Step 3: Evaluate

I_{1}

by parts. Since

\dfrac{d}{dx}e^{\sin x}=\cos x\,e^{\sin x}

, take

u=x,\ dv=\cos x\,e^{\sin x}\,dx\Rightarrow v=e^{\sin x}

I_{1}=x\,e^{\sin x}-\int e^{\sin x}\,dx

Step 4: Evaluate

I_{2}

using the

e

-product rule. Since

\dfrac{d}{dx}\big(\sec x\,e^{\sin x}\big)=\sec x\tan x\,e^{\sin x}+\sec x\cos x\,e^{\sin x}=\sec x\tan x\,e^{\sin x}+e^{\sin x},

integrating gives

I_{2}=\sec x\,e^{\sin x}-\displaystyle\int e^{\sin x}\,dx

. Step 5: Subtract; the

\displaystyle\int e^{\sin x}\,dx

terms cancel:

I=\Big[x\,e^{\sin x}-\int e^{\sin x}\,dx\Big]-\Big[\sec x\,e^{\sin x}-\int e^{\sin x}\,dx\Big]=x\,e^{\sin x}-e^{\sin x}\sec x+C

Correct answer: (3)

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