1. Trig Integration

Hello, the book answer for this is ln2 or .693

They let u be 1+sint and go from there.
I did it another way but am gettin the wrong answer (which is 1)

Thanks

2. Re: Trig Integration

The first step was your mistake. $\frac{\cos t}{1+\sin t} \neq \cos t + \frac{\cos t}{\sin t}$

So you're given $\int_0^\frac{\pi}{2} \frac{\cos t}{1 + \sin t} \, \mathrm{d}x$

It would be better to set u equal to the denominator since the derivative of 1 is 0, allowing you to substitute more.

 Set u as the denominator img.top {vertical-align:15%;} $u = 1 + \sin t$ img.top {vertical-align:15%;} $\mathrm{d}u = \cos t \, \mathrm{d}t$ Substitute img.top {vertical-align:15%;} $\int_0^\frac{\pi}{2} \frac{1}{u} \, \mathrm{d}u$ Integrate img.top {vertical-align:15%;} $\ln u \, \biggr|_0^\frac{\pi}{2}$ Back track u img.top {vertical-align:15%;} $\ln |1 + \sin t \, | \, \biggr|_0^\frac{\pi}{2}$ Evaluate anti-derivative at limits img.top {vertical-align:15%;} $\ln |1 + \sin \frac{\pi}{2} \, | - \ln |1 + \sin 0 |$ Simplify img.top {vertical-align:15%;} $2$

3. Re: Trig Integration

Originally Posted by ReneG
The first step was your mistake. $\frac{\cos t}{1+\sin t} \neq \cos t + \frac{\cos t}{\sin t}$

So you're given $\int_0^\frac{\pi}{2} \frac{\cos t}{1 + \sin t} \, \mathrm{d}x$

It would be better to set u equal to the denominator since the derivative of 1 is 0, allowing you to substitute more.

 Set u as the denominator img.top {vertical-align:15%;} $u = 1 + \sin t$ img.top {vertical-align:15%;} $\mathrm{d}u = \cos t \, \mathrm{d}t$ Substitute img.top {vertical-align:15%;} $\int_0^\frac{\pi}{2} \frac{1}{u} \, \mathrm{d}u$ Integrate img.top {vertical-align:15%;} $\ln u \, \biggr|_0^\frac{\pi}{2}$ Back track u img.top {vertical-align:15%;} $\ln |1 + \sin t \, | \, \biggr|_0^\frac{\pi}{2}$ Evaluate anti-derivative at limits img.top {vertical-align:15%;} $\ln |1 + \sin \frac{\pi}{2} \, | - \ln |1 + \sin 0 |$ Simplify img.top {vertical-align:15%;} $2$

When you make substitution you must change the limits For example
u = 1+sin t
When t = 0 u = 1 and when t = pi/2 , u = 2
Thus the integral would be from 1 to 2, intgreal 1/u du
This will avoid some steps and would help in getting solutin early. in this case it would be simply log 2 since log 1 = 0

4. Re: Trig Integration

Originally Posted by ibdutt
When you make substitution you must change the limits For example
u = 1+sin t
When t = 0 u = 1 and when t = pi/2 , u = 2
Thus the integral would be from 1 to 2, intgreal 1/u du
This will avoid some steps and would help in getting solutin early. in this case it would be simply log 2 since log 1 = 0
Thanks for the heads up. I nice intuitive trick.