# Integral question:

• Mar 25th 2009, 04:12 PM
Kaln0s
Integral question:
Hello, sorry I can't really use the math function well yet I need to learn it :(

Integral (1/t^2)cos(1/t - 1)dt

I substituted with u = 1/t - 1 and du = 1/t^2

For a final answer I get cos(1/t - 1) + c but the book says it's
-sin(1/t - 1) + c.

I don't see when the cosine was derived?..

• Mar 25th 2009, 04:19 PM
gosualite
Quote:

Originally Posted by Kaln0s
Hello, sorry I can't really use the math function well yet I need to learn it :(

Integral (1/t^2)cos(1/t - 1)dt

I substituted with u = 1/t - 1 and du = 1/t^2

For a final answer I get cos(1/t - 1) + c but the book says it's
-sin(1/t - 1) + c.

I don't see when the cosine was derived?..

SO the original problem (I think) was $\displaystyle \int \frac{1}{t^2} \cos\left(\frac{1}{t}-1\right)\,dt$.

Then if you make your substitution, $\displaystyle u = \frac{1}{t}-1$, we have $\displaystyle du = -\frac{1}{t^2}\,dt$.

(Then $\displaystyle dt = -t^2 du$)

Substituting back gives:

$\displaystyle \int \frac{1}{t^2}\cos(u)\,(-t^2)\,du$

reducing to:

$\displaystyle \int -\cos(u)\,du = -\int \cos(u) \,du$.

Since the antiderivative of cosine is sine, this is:

$\displaystyle -\sin(u)+C$.

Now substitute away the u back to $\displaystyle \frac{1}{t}-1$.
• Mar 25th 2009, 04:25 PM
treetheta
Recall the integral of cos is sin