# Thread: Calc 2- Integration by Trigonmetric Substitution

1. ## Calc 2- Integration by Trigonmetric Substitution

I was hoping someone could explain to me how to do the problem

The integral from -pi/2 to 0 of cos(t)/(sqrt(1+sin^2(t)))dt

if its possible will you please show every step because i am really bad at this!!!!

Thank you

2. Hello, fruitkate21!

If you're "really bad at this", this problem is probably fatal . . .

$\int^0_{\text{-}\frac{\pi}{2}}\frac{\cos t\,dt}{\sqrt{1+\sin^2\!t}}$ . . This takes two subsitutions.

Let: $u \:=\:\sin t \quad\Rightarrow\quad du \:=\:\cos t\,dt$

Substitute: . $\int\frac{du}{\sqrt{1+u^2}}$

Let: $u = \tan\theta \quad\Rightarrow\quad du = \sec^2\theta\,d\theta \quad\Rightarrow\quad \sqrt{1+u^2} \:=\:\sec\theta$

Substitute: . $\int\frac{\sec^2\!\theta\,d\theta}{\sec\theta} \;=\;\int\sec\theta\,d\theta \;=\;\ln|\sec\theta + \tan\theta|$

Back-substitute: . $\tan\theta \:=\:\frac{u}{1} \:=\:\frac{opp}{adj} \quad\Rightarrow\quad hyp \:=\:\sqrt{1+u^2} \quad\Rightarrow\quad\sec\theta \:=\:\sqrt{1+u^2}$

. . and we have: . $\ln|u + \sqrt{1+u^2}|$

Back-substitute: . $u = \sin t$

. . and we have: . $\ln\left|\,\sin t + \sqrt{1 + \sin^2\!t}\,\right|\:\bigg]^0_{\text{-}\frac{\pi}{2}}$

I'll let you evaluate it . . .

3. Originally Posted by fruitkate21
I was hoping someone could explain to me how to do the problem

The integral from -pi/2 to 0 of cos(t)/(sqrt(1+sin^2(t)))dt

if its possible will you please show every step because i am really bad at this!!!!

Thank you
First off, if you make the substitution $z=\sin t$, I leave it for you to show that the integral becomes $\int_{-1}^0\frac{\,dz}{\sqrt{1+z^2}}$

Now apply the trig substitution $z=\tan\theta\implies\,dz=\sec^2\theta\,d\theta$ Thus $\int_{-1}^0\frac{\,dz}{\sqrt{1+z^2}}\xrightarrow{z=\tan\t heta}\int_{-\frac{\pi}{4}}^0\frac{\sec^2\theta\,d\theta}{\sqrt {1+\tan^2\theta}}=\int_{-\frac{\pi}{4}}^0\sec\theta\,d\theta$.

Now if you know how to evaluate $\int\sec\theta\,d\theta$, then see my next paragraph. I'll assume for now you don't know how to. Note that $\int\sec\theta\,d\theta=\int\sec\theta\frac{\sec\t heta+\tan\theta}{\sec\theta+\tan\theta}\,d\theta=\ int\frac{\sec^2\theta+\sec\theta\tan\theta}{\sec\t heta+\tan\theta}\,d\theta$. Now let $\eta=\sec\theta+\tan\theta\implies\,d\eta=\left(\s ec\theta\tan\theta+\sec^2\theta\right)\,d\theta$. Thus, the integral becomes $\int\frac{\,d\eta}{\eta}=\ln\!\left|\eta\right|+C= \ln\!\left|\sec\theta+\tan\theta\right|+C$.

Since we were dealing with $\int_{-\frac{\pi}{4}}^0\sec\theta\,d\theta$, we see that this is equivalent to $\left.\bigg[\ln\!\left|\sec\theta+\tan\theta\right|\bigg]\right|_{-\frac{\pi}{4}}^0$. Thus, we have $\ln\!\left|\sec\!\left(0\right)+\tan\!\left(0\righ t)\right|-\ln\!\left|\sec\!\left(-\frac{\pi}{4}\right)+\tan\!\left(-\frac{\pi}{4}\right)\right|=\color{red}\boxed{-\ln\!\left(\sqrt{2}-1\right)}$

Hopefully, you were able to follow all of this. Does it make sense?

EDIT: Soroban beat me to it...