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$$ \int_0^{π/4} (\cos 2x)^{3/2} \cos(x) \, \mathrm{d}x $$

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I was able to find out the correct answer as $\frac{3\pi}{16\sqrt{2}}$ using $t = \frac{\sin\theta}{\sqrt{2}}$ instead of above. But I want to know what mistake I might have done in above method.

Sangchul Lee
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Daksh
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1 Answers1

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At the end of the third line, you have $2t^4 - t^2 +\frac{1}{2}$, which you have written as $(\sqrt{2}t^2- \frac{1}{\sqrt{2}})^2$ in the next line.

But $(\sqrt{2}t^2- \frac{1}{\sqrt{2}})^2=2t^4 - 2t^2 +\frac{1}{2}$

Kendall
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