Integral involving powers of trig functions

[Memo]

Homework Statement

∫(sinx+sin^3x)dx/(cos2x)

Relevant Equations

cos2x=2cos^2x-1

Could you check if my answer is correct? Thank you very much!
Is therea simpler way to solve the math?

 

[PeroK]

Homework Statement: ∫(sinx+sin^3x)dx/(cos2x)
Relevant Equations: cos2x=2cos^2x-1

View attachment 334635
Could you check if my answer is correct?

You can do that yourself by differentiating your answer and checking you get the original integrand.

Thank you very much!
Is therea simpler way to solve the math?

Your method looks good to me. Maybe there's a trick, but not always.

 

[Memo]

You can do that yourself by differentiating your answer and checking you get the original integrand.

Could you tell me how?

 

[Mark44]

You can do that yourself by differentiating your answer and checking you get the original integrand.

Could you tell me how?

If you integrate a function f(x) and get an antiderivative F(x) + C, you can check your answer by differentiating F(x). If your antiderivative is correct, the result will be f(x).

In symbols...
If $\int f(x) dx = F(x) + C$, then $\frac d{dx}\left(F(x) + C\right) = f(x)$

 

[Memo]

If you integrate a function f(x) and get an antiderivative F(x) + C, you can check your answer by differentiating F(x). If your antiderivative is correct, the result will be f(x).

In symbols...
If $\int f(x) dx = F(x) + C$, then $\frac d{dx}\left(F(x) + C\right) = f(x)$

It appears that I was wrong

 

[Gavran]

$\text { The method is good, but there are two mistakes. }$
$\text { The first one is: } u ^ 2 \text { is missed }$
$\text { in the part where } \int \frac { u ^ 2 – 2 } { ( \sqrt 2 u – 1 ) ( \sqrt 2 u + 1 ) } \, du \text { becomes } \int \frac { 1 } { \sqrt 2 u + 1} \, du - \int \frac { 1 } { \sqrt 2 u – 1 } \, du \text { . }$
$\text { The second one is: } \sqrt 2 \text { is missed }$
$\text { in the part where } \int \frac { 1 } { \sqrt 2 u + 1 } \, du - \int \frac { 1 } { \sqrt 2 u – 1 } \ , du \text { becomes } \ln | \sqrt 2 u + 1 | - \ln | \sqrt 2 u - 1 | \text { . }$

$\text { ... and there is not a simpler way. }$

 

[WWGD]

If you integrate a function f(x) and get an antiderivative F(x) + C, you can check your answer by differentiating F(x). If your antiderivative is correct, the result will be f(x).

In symbols...
If $\int f(x) dx = F(x) + C$, then $\frac d{dx}\left(F(x) + C\right) = f(x)$

Look up the Fundamental Theorem of Calculus.

 

[pasmith]

Homework Statement: ∫(sinx+sin^3x)dx/(cos2x)
Relevant Equations: cos2x=2cos^2x-1

View attachment 334635
Could you check if my answer is correct? Thank you very much!
Is therea simpler way to solve the math?

You have correctly obtained $$\int \frac{\sin x + \sin^3 x}{\cos 2x}\,dx = \int \frac{u^2 - 2}{2u^2 - 1}\,du.$$ But you then obtain the partial fraction decomposition of $\frac{1}{2u^2 - 1}$, which is not your integrand; the numerator is $u^2 - 2$ not 1. So you need a further step first: $$\begin{split} \int \frac{u^2 - 2}{2u^2 - 1}\,du &= \frac12 \int \frac{2u^2 - 4}{2u^2 - 1}\,du \\ &= \frac 12 \int 1 - \frac{3}{2u^2 - 1}\,du \\ &= \frac u2 - \frac{3}{4} \int \frac{1}{u^2 - \frac12}\,du\end{split}$$ and now you can use your partial fraction decomposition.