Author Topic: please help with integrating these (Read 709 times) Tweet Share

M-D · « **on:** May 26, 2013, 04:48:13 pm »

Hi

I have 2 questions, i would greatly appreciate if someone could please integrate them and show their steps.

1) integrate (sec(2x))^6 dx
2) integrate (sin(x))^3 dx

thanks a lot

b^3 · « **Reply #1 on:** May 26, 2013, 05:14:18 pm »

1) Try playing around with the identity $\tan^{2}\left(2x\right)+1=\sec^{2}\left(2x\right)$

Spoiler

$\begin{alignedat}{1}\int\sec^{6}\left(2x\right)dx \\ \sin^{2}\left(2x\right)+\cos^{2}\left(2x\right) & =1 \\ \div\cos^{2}\left(2x\right) \\ \tan^{2}\left(2x\right)+1 & =\sec^{2}\left(2x\right) \\ \int\sec^{6}\left(2x\right)dx & =\int\sec^{2}\left(2x\right)\left(\tan^{2}\left(2x\right)+1\right)^{2}dx \\ \text{Let }u=\tan\left(2x\right) \\ \frac{du}{dx}=2\sec^{2}\left(2x\right) \\ \sec^{2}\left(2x\right)dx=\frac{1}{2}du \\ \int\sec^{6}\left(2x\right)dx & =\frac{1}{2}\int\left(u^{2}+1\right)^{2}du \\ & =\frac{1}{2}\int u^{4}+2u^{2}+1du \end{alignedat} $
That should help you complete it.

2). Try playing around with the identity $\sin^{2}(2x)+\cos^{2}(2x)=1$

Spoiler

$\begin{alignedat}{1}\int\sin^{3}\left(x\right)dx & =\int\sin\left(x\right)\sin^{2}\left(x\right)dx \\ & =\int\sin\left(x\right)\left(1-\cos^{2}\left(x\right)\right)dx \\ \text{Let }u=\cos\left(x\right) \\ \frac{du}{dx}=-\sin\left(x\right) \\ \sin\left(x\right)dx=-du \\ \int\sin^{3}\left(x\right)dx & =-\int1-u^{2}du \end{alignedat} $
Again that should help you finish it off

BigAl · « **Reply #2 on:** May 26, 2013, 05:15:01 pm »

Use trig identities and write the integral with sec and tan, sin and cos..then use u sub

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Author Topic: please help with integrating these (Read 709 times) Tweet Share

M-D

please help with integrating these

b^3

Re: please help with integrating these

BigAl

Re: please help with integrating these

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