Integration Techniques

Integration U-substitution - Trigonometric

Evaluate the integral $\displaystyle{ \int { \frac{10x}{\sqrt{3-2x-x^2}} dx }} ,$ using $C$ as the constant of integration.

-10 \sqrt{3-2x-x^2} - 10 \tan^{-1} \left( \frac{x}{2} \right) + C

-10 \sqrt{3-2x-x^2} - 10 \sin^{-1} \left( \frac{x+1}{2} \right) + C

-10 \sqrt{3-2x-x^2} - 10 \cos^{-1} \left( \frac{x+1}{2} \right) + C

-10 \sqrt{3-2x-x^2} - 10 \sin^{-1} \left( \frac{x+3}{2} \right) + C

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by Brilliant Staff

Evaluate the integral $\displaystyle{ \int { \frac { 47 \sqrt{9-x^2} }{x^2 } dx } },$ using $C$ as the constant of integration.

-\frac{ 47 \sqrt{ 9-x^2 } }{x} - 47\tan^{-1} \left( \frac{x}{3} \right) + C

-\frac{ 47 \sqrt{ 9-x^2 } }{x} - 47\sin^{-1} \left( \frac{x}{3} \right) + C

-\frac{ 47 \sqrt{ 9-x^2 } }{x} - 47\cos^{-1} \left( \frac{x}{3} \right) + C

-\frac{ 47 \sqrt{ 9-x^2 } }{x} - 47\cot^{-1} \left( \frac{x}{3} \right) + C

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by Brilliant Staff

Evaluate the integral $\displaystyle{ \int{\frac{15}{x^2\sqrt{x^2+4}}dx}},$ using $C$ as the constant of integration.

-\frac{15 \sqrt{x^2+4} }{ 6x} + C

-\frac{15 \sqrt{x^2+4} }{ 2x} + C

-\frac{15 \sqrt{x^2+4} }{ 4x} + C

-\frac{15 \sqrt{x^2+4} }{ 8x} + C

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Evaluate the integral $\displaystyle{ 43\int { \cos^3 x dx} },$ using $C$ as the constant of integration.

43 \cos x - \frac{43 \sin^3 x}{3} + C

43 \sin x - \frac{43 \sin^3 x}{3} + C

43 \sin x - \frac{43 \cos^3 x}{3} + C

43 \cos x - \frac{43 \cos^3 x}{3} + C

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Given $a>0 ,$ evaluate the integral $\displaystyle{ 41 \int { \frac{dx}{\sqrt{x^2 - a^2}} }},$ using $C$ as the constant of integration.

41 \ln \left \lvert ax^2 + \sqrt{x^2 - a^2} \right \rvert + C

41 \ln \left \lvert ax + \sqrt{x^2 - a^2} \right \rvert + C

41 \ln \left \lvert x^2 + \sqrt{x^2 - a^2} \right \rvert + C

41 \ln \left \lvert x + \sqrt{x^2 - a^2} \right \rvert + C

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by Brilliant Staff