Indefinite IntegrationmediumFree

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

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

Question
∫esin⁡x xcsc⁡x−sec⁡3xcsc⁡x sec⁡x dx\displaystyle\int e^{\sin x}\,\dfrac{x\csc x-\sec^{3}x}{\csc x\,\sec x}\,dx equals
Aesin⁡x−esin⁡xsec⁡x+Ce^{\sin x}-e^{\sin x}\sec x+C
Bx esin⁡x+esin⁡xsec⁡x+Cx\,e^{\sin x}+e^{\sin x}\sec x+C
Cx esin⁡x−esin⁡xsec⁡x+Cx\,e^{\sin x}-e^{\sin x}\sec x+Ccorrect
Dx esin⁡x−sec⁡x+Cx\,e^{\sin x}-\sec x+C
Solution
Step 1: Simplify the integrand by dividing through by csc⁡x sec⁡x\csc x\,\sec x (i.e. multiplying by sin⁡xcos⁡x\sin x\cos x):
xcsc⁡x⋅sin⁡xcos⁡x=xcos⁡x,sec⁡3x⋅sin⁡xcos⁡x=sin⁡xcos⁡2x=tan⁡xsec⁡xx\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
⇒integrand=esin⁡x(xcos⁡x−tan⁡xsec⁡x)\Rightarrow \text{integrand}=e^{\sin x}\big(x\cos x-\tan x\sec x\big)
 Step 2: Split into two integrals:
I=∫xcos⁡x esin⁡x dx−∫tan⁡xsec⁡x esin⁡x dx=I1−I2I=\int x\cos x\,e^{\sin x}\,dx-\int \tan x\sec x\,e^{\sin x}\,dx=I_{1}-I_{2}
 Step 3: Evaluate I1I_{1} by parts. Since ddxesin⁡x=cos⁡x esin⁡x\dfrac{d}{dx}e^{\sin x}=\cos x\,e^{\sin x}, take u=x, dv=cos⁡x esin⁡x dx⇒v=esin⁡xu=x,\ dv=\cos x\,e^{\sin x}\,dx\Rightarrow v=e^{\sin x}:
I1=x esin⁡x−∫esin⁡x dxI_{1}=x\,e^{\sin x}-\int e^{\sin x}\,dx
 Step 4: Evaluate I2I_{2} using the ee-product rule. Since
ddx(sec⁡x esin⁡x)=sec⁡xtan⁡x esin⁡x+sec⁡xcos⁡x esin⁡x=sec⁡xtan⁡x esin⁡x+esin⁡x,\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 I2=sec⁡x esin⁡x−∫esin⁡x dxI_{2}=\sec x\,e^{\sin x}-\displaystyle\int e^{\sin x}\,dx. Step 5: Subtract; the ∫esin⁡x dx\displaystyle\int e^{\sin x}\,dx terms cancel:
I=[x esin⁡x−∫esin⁡x dx]−[sec⁡x esin⁡x−∫esin⁡x dx]=x esin⁡x−esin⁡xsec⁡x+CI=\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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