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Integrate ∫(1+cos4x)/(cotx – tanx) dx

Integrate
∫(1+cos4x)/(cotx – tanx) dx

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int \frac{1+cos4x}{cotx - tanx}dx
I = \int \frac{1+cos4x}{\frac{cosx}{sinx} - \frac{sinx}{cosx}}dx
I = \int \frac{(1+cos4x).sinx.cosx}{cos^{2}x-sin^{2}x}dx
I = \int \frac{(2cos^{2}2x).sinx.cosx}{cos^{2}x-sin^{2}x}dx
I = \int \frac{(cos^{2}2x).sin2x}{cos{2}x}dx
I = \int cos2x.sin2xdx
I = \frac{1}{2}\int 2cos2x.sin2xdx
I = \frac{1}{2}\int sin4xdx
I = \frac{-cos4x}{8} + c

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