Getting wrong answer trying to evaluate $\int \frac {\sin(2x)dx}{(1+\cos(2x))^2}$

Question

I'm trying to evaluate $$\int \frac {\sin(2x)dx}{(1+\cos(2x))^2} = I$$

Here's what I've got:

$$t=1+\cos(2x)$$ $$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$ $$I = \int \frac 1{t^2} \cdot \frac {-1}2 dt = -\frac 12 \int t^{-2}dt = -\frac 12 \frac {t^{-2+1}}{-2+1}+C$$ $$=-\frac 12 \frac {t^{-1}}{-1}+C = \frac 1{2t} +C = \frac 1{2(1+\cos(2x)}+C$$

I need to find what went wrong with my solution. I know the answer is correct but the line $dt=-2\sin(2x)dx$ concludes that $dx=\frac {dt}{-2}$ is false. But why? What should it be instead?

Answer 1

Probably, you saw the solution as $$\frac{\sec^2(x)}{4}+C$$

But note that $$\begin{array}{rclcc} \frac{1}{2}\frac{1}{1+\cos(2x)}&=&\frac{1}{2}\frac{1}{1+\cos^2(x)-\sin^2(x)}&/\,\,\cos(2x)=\cos^2(x)-\sin^2(x)\\ &=&\frac{1}{2}\frac{1}{1+\cos^2(x)-\sin^2(x)}&/\,\,1=\cos^2(x)+\sin^2(x)\\ &=&\frac{1}{2}\frac{1}{2\cos^2(x)}\\ &=&\frac{\sec^2(x)}{4}\end{array}$$

Answer 2

the derivative of your function is given by $$\frac{1}{2}\left(\left(\left(1+\cos(2x)\right)\right)^{-1}\right)'$$$$=-\frac{1}{2}\left(1+\cos(2x)\right)^{-2}(-\sin(2x)\cdot 2)$$ this is your function

Answer 3

Perhaps simpler:

$$\begin{cases}1+\cos2x=1+2\cos^2x-1=2\cos^2x\\{}\\\sin2x=2\sin x\cos x\end{cases}\;\;\;\;\;\;\;\;\;\;\;\implies$$

$$\int\frac{\sin2x}{(1+\cos2x)^2}=\int\frac{2\sin x\cos x}{4\cos^4x}dx=-\frac12\int\frac{(\cos x)'dx}{\cos^3x}=-\frac12\frac{\cos^{-2}x}{-2}+C=$$

$$=\frac14\sec^2x+C\ldots\ldots\text{and your solution is correct, of course}$$

using that

$$\int\frac1{x^3}dx=-\frac12\frac1{x^2}+c\implies \int\frac{f'(x)}{f(x)^3}dx=-\frac1{2f(x)^2}+c$$

Answer 4

Your solution is good. The substitution $t=1+\cos2x$ gives $$ dt=-2\sin2x\,dx $$ so the integral becomes $$ \int\frac{\sin2x}{(1+\cos2x)^2}\,dx=\int-\frac{1}{2}\frac{1}{t^2}\,dt=\frac{1}{2t}+C=\frac{1}{2(1+\cos2x)}+C $$ Differentiating shows the correctness. However, there's no “canonical” answer; if you notice that $1+\cos2x=2\cos^2x$ you can as well rewrite the answer as $$ \frac{1}{4\cos^2x}+C $$ but also as $$ \frac{1+\tan^2x}{4}+C $$ or infinitely many variations thereof.

Answer 5

Your only error was where you said

$$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$

You should have said

$$dt=-2\sin(2x)dx \implies \sin(2x)dx = \frac {dt}{-2}$$

You need the $\sin(2x)$ because the original integral has it in the numerator:

$${\sin(2x)dx\over(1+\cos(2x))^2}={1\over(1+\cos(2x))^2}(\sin(2x)dx)={1\over t^2}\cdot{dt\over-2}$$

Everything else you did was fine, if perhaps a little overly elaborated.

Answer 6

$$\int \frac {\sin(2x)dx}{(1+\cos(2x))^2} = [1+cos2x = u, -2sin2x dx = du] = -0.5 \int \frac {du}{u^2}= \frac{0.5}{u}+c=\frac{1}{2+2cos2x}+c $$