Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\int x\times\sqrt{8-x^2} \,dx = \,?$$ I got to this: $$\int\sqrt{8x^2-x^4} \,dx$$ or: $$\int\frac{8x-x^3}{\sqrt{8+x^2}}\, dx$$

I don't know how to integrate neither. If possible, no $\sin$ \ $\cos$ \ etc.

share|improve this question
1  
Have you seen the method of substitution before, sometimes called $u$-substitution? –  Graphth Nov 3 '12 at 20:11

2 Answers 2

up vote 5 down vote accepted

Make the substitution $u=8-x^2$, and you’ll get an easy integration.

share|improve this answer

Or directly:

$$\int x\sqrt{8-x^2}\,dx=-\frac{1}{2}\int\sqrt{8-x^2}\,d(8-x^2)=-\frac{1}{2}\frac{2}{3}(8-x^2)^{3/2}+C$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.