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FP3 integration

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Hi guys

Can anyone help me with Question 12? a? I tried what the question suggested but didn't know what to do next.

any reply will be appreciated. Thanks


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Avatar for Zacken
Zacken
22
Original post by 123321123

Hi guys

Can anyone help me with Question 12? a? I tried what the question suggested but didn't know what to do next.

any reply will be appreciated. Thanks


Posted from TSR Mobile


IBP with tan⁡nx=tan⁡2xtan⁡n−2x=(sec⁡2x−1)tan⁡n−2x=sec⁡2tan⁡n−2x−tan⁡n−2x\tan^n x = \tan^2 x \tan^{n-2} x = (\sec^2 x - 1)\tan^{n-2} x = \sec^2 \tan^{n-2} x - \tan^{n-2}x

So ∫tan⁡nx dx=[tan⁡n−1xn−1]0π/4−In−2\displaystyle \int \tan^n x \, \mathrm{d}x = \bigg[\frac{\tan^{n-1} x}{n-1} \bigg]_0^{\pi/4} - I_{n-2}

By using the reverse chain rule on ∫sec⁡2xtan⁡n−2x dx\int \sec^2 x \tan^{n-2} x \, \mathrm{d}x.
(edited 8 years ago)
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123321123
OP
2
Original post by Zacken
IBP with tan⁡nx=tan⁡2xtan⁡n−2x=(sec⁡2x−1)tan⁡n−2x=sec⁡2tan⁡n−2x−tan⁡n−2x\tan^n x = \tan^2 x \tan^{n-2} x = (\sec^2 x - 1)\tan^{n-2} x = \sec^2 \tan^{n-2} x - \tan^{n-2}x

So ∫tan⁡nx dx=[tan⁡n−1xn−1]0π/4−In−2\displaystyle \int \tan^n x \, \mathrm{d}x = \bigg[\frac{\tan^{n-1} x}{n-1} \bigg]_0^{\pi/4} - I_{n-2}

By using the reverse chain rule on ∫sec⁡2xtan⁡n−2x dx\int \sec^2 x \tan^{n-2} x \, \mathrm{d}x.


thanks so much, i understand it now.
Reply 3
Avatar for Zacken
Zacken
22
Original post by 123321123
thanks so much, i understand it now.


Cool.

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