MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Special Integrals - 3

solve int;...

Question

solve
∫cos⁡θ/sin⁡4θ dθ

Open in App

Solution

Dear Student,

$\int \frac{\cos (θ)}{\sin (4 θ)} d θ = \int \frac{\cos (θ)}{4 \sin (θ) \cos^{3} (θ) - 4 \sin^{3} (θ) \cos (θ)} d θ = \frac{1}{4} \int \frac{\cos (θ)}{\sin (θ) \cos (θ) [\cos^{2} (θ) - \sin^{2} (θ)]} d θ = \frac{1}{4} \int \frac{1}{\sin (θ) [\cos^{2} (θ) - (1 - \cos^{2} (θ))]} d θ = \frac{1}{4} \int \frac{\sin (θ)}{\sin^{2} (θ) [2 \cos^{2} (θ) - 1]} d θ = \frac{1}{4} \int \frac{- \sin (θ)}{(\cos^{2} (θ) - 1) [2 \cos^{2} (θ) - 1]} d θ L e t \cos (θ) = u \Rightarrow - \sin (θ) d θ = d u \Rightarrow \int \frac{\cos (θ)}{\sin (4 θ)} d θ = \frac{1}{4} \int \frac{1}{(u^{2} - 1) [2 u^{2} - 1]} d u = \frac{1}{4} \int \frac{1}{(u + 1) (u - 1) (2 u - \sqrt{2}) (2 u + \sqrt{2})} d u = \frac{1}{4} \int \frac{\sqrt{2}}{(2 u + \sqrt{2})} - \frac{\sqrt{2}}{(2 u - \sqrt{2})} - \frac{1}{2 (u + 1)} + \frac{1}{2 (u - 1)} d u = \frac{1}{4} [\sqrt{2} \int \frac{d u}{(2 u + \sqrt{2})} - \sqrt{2} \int \frac{d u}{(2 u - \sqrt{2})} - \frac{1}{2} \int \frac{d u}{(u + 1)} + \frac{1}{2} \int \frac{d u}{(u - 1)}] = \frac{1}{4} [\frac{\sqrt{2} \ln (2 u + \sqrt{2})}{2} - \frac{\sqrt{2} \ln (2 u - \sqrt{2})}{2} - \frac{\ln (u + 1)}{2} + \frac{\ln (u - 1)}{2}] + C = \frac{1}{4} [\frac{\ln (2 u + \sqrt{2})}{\sqrt{2}} - \frac{\sqrt{2} \ln (2 u - \sqrt{2})}{\sqrt{2}} - \frac{\ln (u + 1)}{2} + \frac{\ln (u - 1)}{2}] + C = \frac{1}{4} [\frac{\ln (2 \cos (θ) + \sqrt{2})}{\sqrt{2}} - \frac{\sqrt{2} \ln (2 \cos (θ) - \sqrt{2})}{\sqrt{2}} - \frac{\ln (\cos (θ) + 1)}{2} + \frac{\ln (\cos (θ) - 1)}{2}] + C$
Regards,

Suggest Corrections

thumbs-up

0

Similar questions

\int \frac{sin 4 x}{sin x} d x

\frac{d}{d x} [sin 4 x - cos 4 x + tan \sqrt{e^{x}}]

(1−cosθ+sinθ)(1+cosθ+sinθ)=2sinθ(1−cos⁡θ+sin⁡θ)(1+cos⁡θ+sin⁡θ)=2sin⁡θ

\int {sin}^{4} x

\int \frac{{sin}^{4} x}{{cos}^{2} x} d x

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Special Integrals - 3

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Special Integrals - 3

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon