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Question Number 31951 by NECx last updated on 17/Mar/18

$$Evaluate\int \sin \sqrt{x}dx$$

Answered by mrW2 last updated on 17/Mar/18

$$u=\sqrt{x}$$$$du=\frac{dx}{2\sqrt{x}}=\frac{dx}{2u}$$$$dx=2udu$$$$\int \sin \sqrt{x}dx$$$$=\int 2u\sin udu$$$$=-2\int ud\cos u$$$$=-2[u\cos u-\int \cos udu]$$$$=-2[u\cos u-\sin u]+C$$$$=2[\sin u-u\cos u]+C$$$$=2[\sin \sqrt{x}-\sqrt{x}\cos \sqrt{x}]+C$$

Commented by mondodotto@gmail.com last updated on 19/Mar/18

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

Commented by NECx last updated on 20/Mar/18

$$thankssomuch!$$

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