integrals using trig substitution

[kidmo87]

Hello everyone. Im having issues leading to the answer. I know the answer is 1/6 tan theta + c, but im stuck on how to get there, any advice would be appreciated. thanks.

 

[srmichael]

Hello everyone. Im having issues leading to the answer. I know the answer is 1/6 tan theta + c, but im stuck on how to get there, any advice would be appreciated. thanks.View attachment 2720

Remember, 1 - sin²Θ = cos²Θ, so the denominator becomes 64(cosΘ)^3 and then one of these cancels with the cosΘ in your du. Can you take it from here?

 

[kidmo87]

so for the numerator, instead of one, I should put 4 cos theta?

 

[srmichael]

so for the numerator, instead of one, I should put 4 cos theta?

Yes. You should end up with:

\(\displaystyle \int\frac{4\cos\theta}{64\cos^3\theta}\ d \theta\)

which simplifies to:

\(\displaystyle \int\frac{1}{16\cos^2\theta}\ d \theta\)

Can you take it from here?

Hint: what is an identity for \(\displaystyle \frac{1}{\cos^2\theta}\)

 

[kidmo87]

Yup I got it, thanks a lot!!!