Integral involving powers of trig functions

Memo
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Homework Statement
∫(sinx+sin^3x)dx/(cos2x)
Relevant Equations
cos2x=2cos^2x-1
368064999_867353445000190_1304311522445404453_n.jpg

Could you check if my answer is correct? Thank you very much!
Is therea simpler way to solve the math?
 
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Memo said:
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.
Memo said:
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.
 
PeroK said:
You can do that yourself by differentiating your answer and checking you get the original integrand.
Could you tell me how?
 
PeroK said:
You can do that yourself by differentiating your answer and checking you get the original integrand.

Memo said:
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)##
 
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Mark44 said:
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
 
## \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. } ##
 
Mark44 said:
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.
 
Memo said:
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 <br /> \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}<br /> \int \frac{u^2 - 2}{2u^2 - 1}\,du &amp;= \frac12 \int \frac{2u^2 - 4}{2u^2 - 1}\,du \\<br /> &amp;= \frac 12 \int 1 - \frac{3}{2u^2 - 1}\,du \\<br /> &amp;= \frac u2 - \frac{3}{4} \int \frac{1}{u^2 - \frac12}\,du\end{split} and now you can use your partial fraction decomposition.
 
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