Question

just verify if I am right, please... INTEGRATION
(haven't learned by parts method, don't think it is needed tho)

Answer 1

\[\LARGE \int\limits_{ }^{ }\frac{\sin(2x)}{1+\cos^2(x)}~dx\]

Answer 2

\[u=1+\cos^2(x)\]\[\frac{du}{dx}(1+\cos^2(x))=\frac{du}{dx}(\cos~x)^2=-2\cos(x)\sin(x)=-\sin(2x)\]\[-du=\sin(2x)~dx\]

Answer 3

\[-\LARGE \int\limits_{ }^{ }\frac{1}{u}~dx\]
\[answer,~~~\LARGE -\ln(1+\cos^2x)+C\]

Answer 4

This is the first problem, and the second one is....

Answer 5

\[\LARGE \int\limits_{ }^{ }\frac{\sin(x)}{1+\cos^2(x)}~dx\]This is where I am a little bit stuck

Answer 6

Anyone that can help me?
@mathstudent55 can you give me a suggestion?

Answer 7

No, I got it....
\[u=\cos(x),~-du=\sin(x)\]
\[-\LARGE \int\limits_{ }^{ }\frac{1}{1+u^2}~du\]
-tan^(-1) (cos x) + C

Answer 8

got it, no need helpin me,