forgot how to integrate this (1 Viewer)

wahbetemaujkardi · Jun 8, 2022

test on friday so need help asap if possible

cossine · Jun 8, 2022

wahbetemaujkardi said:
View attachment 35757
test on friday so need help asap if possible

Use an online integral calculator with full steps. My guess is let u = 1+x

stupid_girl · Jun 8, 2022

Let u=sqrt(x)
u^2=x
2u du=dx

Drongoski · Jun 8, 2022

$\int ^1 _0 \frac {\sqrt x}{1 + x} dx = \int ^1 _0 \frac {\sqrt x}{1+(\sqrt x)^2} \times 2\sqrt x d\sqrt x\\ \\ =2\int^1 _0 (1 - \frac {1}{1 + (\sqrt x)^2} )d \sqrt x \\ \\ =2 \left[\sqrt x - tan^{-1} \sqrt x \right ]^1 _0 \\ \\ = 2[(1 - \frac {\pi}{4} ) -(0 - 0)] = 2 - \frac {\pi}{2} \\ \\ ?????$

Ordinary substitution method more straightforward.

Akshara Patil · Jun 8, 2022

stupid_girl said:
Let u=sqrt(x)
u^2=x
2u du=dx

I did it this way
- then split up integral...

yanujw · Jun 8, 2022

Alternatively, let x = tan(theta)^2
You'll end up with the integral of 2tan(theta)^2 with new boundaries pi/4 and 0. I've checked and this confirms Drongoski's answer.

Run hard@thehsc · Jun 8, 2022

yanujw said:
Alternatively, let x = tan(theta)^2
You'll end up with the integral of 2tan(theta)^2 with new boundaries pi/4 and 0. I've checked and this confirms Drongoski's answer.

Yh I used that and it works

wahbetemaujkardi · Jun 10, 2022

cossine said:
Use an online integral calculator with full steps. My guess is let u = 1+x

totally forgot those existed thanks

forgot how to integrate this (1 Viewer)

wahbetemaujkardi

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