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I have trouble calculating calculating this integral: $ \int \cos(x) \sqrt{1+\sin^2(x)} dx$.

I tried to use subsitution $t=\sqrt{1+\sin^2(x)}$ and got $\int \frac{(2-t^2)\sqrt{t^2-1}}{t} dt$ but I'm not sure it's correct and I don't know how to proceed from there.

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  • $\begingroup$ Choose $t=\sin(x)\implies dt=\cos(x)dx$ $\endgroup$
    – Holo
    Commented Jan 14, 2018 at 13:04

2 Answers 2

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Note that in substituting $\sin x = t$, we have $\cos x\, dx = dt$, giving us: $$I = \int \cos x \sqrt{1+\sin^2 x} \, dx $$ $$= \int \sqrt{1+t^2}\, dt$$ which is a standard result.

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with $t=\sin(x)$ you will get $$dt=\cos(x)dx$$ the result should be $$1/2\,\sin \left( x \right) \sqrt {1+ \left( \sin \left( x \right) \right) ^{2}}+1/2\,{\rm arcsinh} \left(\sin \left( x \right) \right)+C $$

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  • $\begingroup$ thank you for your hint, it is corrected now! $\endgroup$
    – Dr. Sonnhard Graubner
    Commented Jan 14, 2018 at 13:14

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