# Thread: [SOLVED] Evaluate the Indefinite Integral

1. ## [SOLVED] Evaluate the Indefinite Integral

Hi,
I'm I on the right track with this problem?

1. The problem statement, all variables and given/known data
Evaluate the Indefinite Integral

2. Relevant equations
integral of sin t sec^2 (cos t)dt

3. The attempt at a solution
integral of sin t sec^2 (cos t)dt

Let u = cos t

du/dt = sin t

dt * -sin t = du

dt = du/-sin t

integral sin t sec^2 (u) dt

integral sin t sec^2 (u) du/-sin t

- integral of sec^2 (u) du

- [tan (u) du ] + c

- tan (cos t) + c

2. Originally Posted by sparky
Hi,
I'm I on the right track with this problem?

1. The problem statement, all variables and given/known data
Evaluate the Indefinite Integral

2. Relevant equations
integral of sin t sec^2 (cos t)dt

3. The attempt at a solution
integral of sin t sec^2 (cos t)dt

Let u = cos t

du/dt = -sin t

dt * -sin t = du

dt = du/-sin t

integral sin t sec^2 (u) dt

integral sin t sec^2 (u) du/-sin t

- integral of sec^2 (u) du

- [tan (u) du ] + c

- tan (cos t) + c
All correct except for the line with red color. It must be a typo! $\displaystyle \frac{d (cost)}{dt}= -sint$

3. Originally Posted by sparky
Hi,
I'm I on the right track with this problem?

1. The problem statement, all variables and given/known data
Evaluate the Indefinite Integral

2. Relevant equations
integral of sin t sec^2 (cos t)dt

3. The attempt at a solution
integral of sin t sec^2 (cos t)dt

Let u = cos t

du/dt = sin t

dt * -sin t = du

dt = du/-sin t

integral sin t sec^2 (u) dt

integral sin t sec^2 (u) du/-sin t

- integral of sec^2 (u) du

- [tan (u) du ] + c

- tan (cos t) + c
differentiate your result to check the work.

4. Originally Posted by harish21
All correct except for the line with red color. It must be a typo! $\displaystyle \frac{d (cost)}{dt}= -sint$
Ok, thanks very much skeeter and harish21