Find $\int \cos(x) \sqrt{1+\sin^2(x)} dx$

by user509051   Last Updated January 14, 2018 13:20 PM

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

Dr. Sonnhard Graubner
January 14, 2018 13:01 PM

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.

Rohan
January 14, 2018 13:02 PM