Question

Calculus
7 * integral from 0 to pi/2 (sec^3 (theta/2) d theta)

Answer 1

\[7 \int\limits_{0}^{\frac{\pi}{2}}\sec^3(\frac{\theta}{2}) d \theta\]

Answer 2

well to integrate sec^3(x) you need integration by parts

Answer 3

haha same pace

Answer 4

if you want to do a substitution you can you know before the integration by parts

Answer 5

let u=theta/2
du=1/ dtheta

Answer 6

oops typed too fast 2 got left out

Answer 7

let u=theta/2
du=1/2 dtheta

Answer 8

and then 2 du=d theta

Answer 9

so we have this?
\[\int\limits_{}^{} udv = \tan(u)\sec(u) - \int\limits_{}^{} tanu*\frac{ 1 }{ 2 }\sec(u)\tan(u) \]

Answer 10

well what exactly is u and and dv
are you not including the constant multiple 7 and just the new constant multiple 2
we can ignore all constant multiples til the end

Answer 11

I would integrate this first:
\[\int\limits_{}^{}\sec^3(u) du\]

Answer 12

IBP is amazing, but if you love partial fractions :
\[\large \sec^3x = \frac{\cos x}{\cos^4x} = \frac{\cos x}{(1-\sin^2x)^2}\]

Answer 13

That is kinda cute.

Answer 14

IBP is more even more cute :
\[\large \begin{align} I = \int\sec^3(u) du &= \int\sec u ~d(\tan u)\\~\\&= \sec u \tan u - \int (\sec u \tan u )\tan u ~du \\~\\&=\sec u \tan u - \int \sec u (\sec^2u - 1) du \\~\\ &= \sec u \tan u - I + \int \sec u ~du\end{align}\]

Answer 15

why the -1 and what is I?

Answer 16

\(\large I = \int \sec^3 u ~du\)

Answer 17

its the value of the integral you're after

Answer 18

Then what?

Answer 19

I'm more lost now than I was before I asked the question. Have I been intentionally mislead?

Answer 20

I don't think anyone misled you here.

Answer 21

What are you confused about?

Answer 22

\[ \\ \int\limits_{0}^{\frac{\pi}{2}}7 \sec^3( \frac{ \theta}{2}) d \theta \\ u=\frac{ \theta}{2} \\ 2u =\theta \\ 2 du= d \theta \\ \text{ If } \theta=0 \text{ then } u=\frac{0}{2}=0 \\ \text{ If } \theta=\frac{\pi}{2} \text{ then } u=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4} \\ \int\limits_{0}^{\frac{\pi}{4}} 7 \sec^3(u) 2 du \\ 14 \int\limits_{0}^{\frac{\pi}{4}} \sec^3(u) du \]
Gan alraedy got you started in finding the general antiderivative for sec^3(u)

Answer 23

you can then multiply that by 14 when you are done solving that equation he has for you

Answer 24

then you can input the limits