Indefinite IntegrationmediumFree

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

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

Question

\displaystyle\int \dfrac{\cos^{3}x+\cos^{5}x}{\sin^{2}x+\sin^{4}x}\,dx

equals

\sin x-6\tan^{-1}(\sin x)+c

\sin x-2(\sin x)^{-1}+c

\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+c

correct

\sin x-2(\sin x)^{-1}+5\tan^{-1}(\sin x)+c

Solution

Step 1: Substitute

t=\sin x\Rightarrow dt=\cos x\,dx

. Pull one

\cos x

from the numerator to supply

dt

, and use

\cos^{2}x=1-t^{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^{2}x+\sin^{4}x=t^{2}+t^{4}

Step 2: The integral becomes

I=\int\dfrac{(1-t^{2})+(1-t^{2})^{2}}{t^{2}+t^{4}}\,dt

Step 3: Expand the numerator:

(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

\dfrac{t^{4}-3t^{2}+2}{t^{2}(t^{2}+1)}

. Dividing, the polynomial part is

1

, leaving

\dfrac{-4t^{2}+2}{t^{2}(t^{2}+1)}=\dfrac{A}{t^{2}}+\dfrac{B}{t^{2}+1}

. Solving

-4t^{2}+2=A(t^{2}+1)+Bt^{2}

gives

A=2

(at

t=0

) and

B=-6

(coefficient of

t^{2}

). So

\text{integrand}=1+\dfrac{2}{t^{2}}-\dfrac{6}{t^{2}+1}

Step 5: Integrate and back-substitute

t=\sin x

I=\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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