Integrate \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}.

Category: QuestionsIntegrate \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}.
Editor">Editor Staff asked 11 months ago

Integrate \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}.

(a) -log⁡(1+2sin⁡2x)+C

(b) \frac{1}{4} log⁡(1-sin⁡2x)+C

(c) –\frac{1}{4} log⁡(1+cos⁡2x)+C

(d) -log⁡(1-sin⁡2x)+C

The question was asked in my homework.

This question is from Methods of Integration-2 in portion Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

Right choice is (d) -log⁡(1-sin⁡2x)+C

The explanation is: \int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=\int \frac{2 cos⁡2x}{cos^2⁡x+sin^2⁡x-sin⁡2x} \,dx \,(∵2 cos⁡x sin⁡x=sin⁡2x)

=\int \frac{2 cos⁡2x}{1-sin⁡2x} dx

Let 1-sin⁡2x=t

Differentiating w.r.t x, we get

-2 cos⁡2x dx=dt

2 cos⁡2x dx=-dt

\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=-\int \frac{dt}{t}

=-log⁡t

Replacing t with 1-sin⁡2x, we get

∴\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=-log⁡(1-sin⁡2x)+C