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Integral of (cos³x+cos⁵x)/(sin²x+sin⁴x) | JEE

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

Question
∫cos⁡3x+cos⁡5xsin⁡2x+sin⁡4x dx\displaystyle\int \dfrac{\cos^{3}x+\cos^{5}x}{\sin^{2}x+\sin^{4}x}\,dx equals
Asin⁡x−6tan⁡−1(sin⁡x)+c\sin x-6\tan^{-1}(\sin x)+c
Bsin⁡x−2(sin⁡x)−1+c\sin x-2(\sin x)^{-1}+c
Csin⁡x−2(sin⁡x)−1−6tan⁡−1(sin⁡x)+c\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+ccorrect
Dsin⁡x−2(sin⁡x)−1+5tan⁡−1(sin⁡x)+c\sin x-2(\sin x)^{-1}+5\tan^{-1}(\sin x)+c
Solution
Step 1: Substitute t=sin⁡x⇒dt=cos⁡x dxt=\sin x\Rightarrow dt=\cos x\,dx. Pull one cos⁡x\cos x from the numerator to supply dtdt, and use cos⁡2x=1−t2\cos^{2}x=1-t^{2}:
cos⁡3x+cos⁡5x=cos⁡x(cos⁡2x+cos⁡4x)=cos⁡x[(1−t2)+(1−t2)2]\cos^{3}x+\cos^{5}x=\cos x\big(\cos^{2}x+\cos^{4}x\big)=\cos x\big[(1-t^{2})+(1-t^{2})^{2}\big]
sin⁡2x+sin⁡4x=t2+t4\sin^{2}x+\sin^{4}x=t^{2}+t^{4}
 Step 2: The integral becomes
I=∫(1−t2)+(1−t2)2t2+t4 dtI=\int\dfrac{(1-t^{2})+(1-t^{2})^{2}}{t^{2}+t^{4}}\,dt
 Step 3: Expand the numerator:
(1−t2)+(1−2t2+t4)=t4−3t2+2,t2+t4=t2(t2+1)(1-t^{2})+(1-2t^{2}+t^{4})=t^{4}-3t^{2}+2,\qquad t^{2}+t^{4}=t^{2}(t^{2}+1)
 Step 4: Decompose t4−3t2+2t2(t2+1)\dfrac{t^{4}-3t^{2}+2}{t^{2}(t^{2}+1)}. Dividing, the polynomial part is 11, leaving −4t2+2t2(t2+1)=At2+Bt2+1\dfrac{-4t^{2}+2}{t^{2}(t^{2}+1)}=\dfrac{A}{t^{2}}+\dfrac{B}{t^{2}+1}. Solving −4t2+2=A(t2+1)+Bt2-4t^{2}+2=A(t^{2}+1)+Bt^{2} gives A=2A=2 (at t=0t=0) and B=−6B=-6 (coefficient of t2t^{2}). So
integrand=1+2t2−6t2+1\text{integrand}=1+\dfrac{2}{t^{2}}-\dfrac{6}{t^{2}+1}
 Step 5: Integrate and back-substitute t=sin⁡xt=\sin x:
I=∫ ⁣(1+2t2−6t2+1)dt=t−2t−6tan⁡−1t+c=sin⁡x−2(sin⁡x)−1−6tan⁡−1(sin⁡x)+cI=\int\!\left(1+\dfrac{2}{t^{2}}-\dfrac{6}{t^{2}+1}\right)dt=t-\dfrac{2}{t}-6\tan^{-1}t+c=\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+c
 Correct answer: (3)
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