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Question Number 192349 by 073 last updated on 15/May/23

Answered by mehdee42 last updated on 15/May/23

$$I=\int \frac{{cos}^{4}x\sqrt{1+sinx}}{cosx}dx=\int (1-{sin}^{2}x)\sqrt{1+sinx}cosxdx$$$$let:1+sinx=u^2\Rightarrow cosxdx=2udu$$$$\Rightarrow I=2\int (-u^2+2u)u^{2}du=2(-\frac{1}{5}{u}^5+\frac{1}{2}{u}^4)+c$$$$\Rightarrow I={(1+sinx)}^2(\frac{1}{2}-\frac{2\sqrt{1+sinx}}{5})+c$$

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