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.

$$\int{\frac{3}{5y^2 + 4}}dy$$

$$\frac{3}{4}\int{\frac{1}{\left(\frac{\sqrt{5}y}{2}\right)^2 + 1}}dy$$

$$u = \frac{\sqrt{5}y}{2}$$

$$dy = \frac{2}{\sqrt{5}}du$$

My solution to this problem was

$$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan\left(\frac{\sqrt{5}y}{2}\right)} + c\right)$$

However, apparently the solution is as follows: $$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan\frac{\sqrt{5}y}{2}}\right) + c $$

My question is: why is c outside the bracket? I was taught that when you integrate a value you must add c. But since we're adding c, we also must multiply by whatever is outside of the integral.

share|improve this question
    
It is just the same. Just expand your expression and you have another constant. –  Claude Leibovici Mar 14 at 6:28
    
Did anybody notice that the integral is wrong? –  alex Mar 14 at 7:11

3 Answers 3

up vote 1 down vote accepted

Remember that $c$ is a generic constant - it doesn't hold a particular value until it is assigned one. In this case, note that $\frac{3}{2\sqrt{5}}*c$ is still a constant. $$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan(\frac{\sqrt{5}y}{2})} + c\right)=\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan(\frac{\sqrt{5}y}{2})}\right)+\frac{3}{2\sqrt{5}}*c$$ $$\frac{3}{2\sqrt{5}}\left(\frac{1}{\tan(\frac{\sqrt{5}y}{2})}\right)+c$$

share|improve this answer

Doesn't matter if c is inside or outside the bracket. It just denotes a constant value. If you multiply the coefficient with the constant, you will just get another constant.

share|improve this answer

This is because what ever the value is that is being multiplied is still a constant. The mathematician does not concern themselves with any magnitude of this constant when solving empirically.

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.