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$$\int_0^{\pi/4} \sec^4 \theta \tan^4 \theta\; d\theta$$

I used the substitution: let $u = \tan \theta$ ... then $du = \sec^2 \theta \; d\theta$.

I know that now I have to change the limits of integration, but am stuck as to how I should proceed.

Should I sub the original limits into $\tan \theta$ or should I let $\tan \theta$ equal the original limits and then get the new limits?

And if it help, the answers of the definite integral is supposed to be $0$.

Thanks in advance.

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\begin{align} I &= \int_{0}^{\pi/4} \sec^{4}(\theta) \tan^{4}(\theta) \ d\theta \\ &= \int_{0}^{\pi/4} \sec^{2}(\theta) ( 1 + \tan^{2}(\theta)) \tan^{4}(\theta) \ d\theta \\ &= \int_{0}^{\pi/4} \tan^{4}(\theta) \ d(\tan\theta) + \int_{0}^{\pi/4} \tan^{6}(\theta) \ d(\tan\theta) \\ &= \left[ \frac{1}{5} \tan^{5}(\theta) \right]_{0}^{\pi/4} + \left[ \frac{1}{7} \tan^{7}(\theta) \right]_{0}^{\pi/4} \\ &= \frac{1}{5} + \frac{1}{7} = \frac{12}{35}. \end{align}

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$\sec^4\theta = (1 + \tan^2\theta)\cdot \sec^2\theta$, then substitute $u = \tan\theta$ to get:

$$I = \displaystyle \int_{0}^1 (u^6 + u^4) du = \left.\dfrac{u^7}{7} + \dfrac{u^5}{5}\right|_{0}^1 = \dfrac{12}{35}.$$

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  • $\begingroup$ +1. I also edit your answer so it looks better, I hope you don't mind. $\endgroup$ – Tunk-Fey May 18 '14 at 2:25
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    $\begingroup$ @Tunk-Fey: feel free. $\endgroup$ – DeepSea May 18 '14 at 2:26

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