Question

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

Answer 1

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