integration (1 Viewer)

nfreidman · Apr 28, 2012

how do you integate

sqrt (x)/1+x as a definite integral between 1 and zero, and with the substition x=tan^2(theta)

thank you!

nightweaver066 · Apr 28, 2012

$\int_0^1 \frac{\sqrt{x}}{1 + x} dx$

$Let ~ x = tan^2\theta$

$dx = 2tan\theta sec^2\theta d\theta$

$x = 1, \theta = \frac{\pi}{4}$

$x = 0, \theta = 0$

$\therefore \int ^{\frac{\pi}{4}}_0 \frac{tan\theta}{1 + tan^2\theta} \times 2tan\theta sec^2\theta d\theta$

$\int^{\frac{\pi}{4}}_0 \frac{2tan^2\theta sec^2\theta}{sec^2\theta}d\theta$

$= \int^{\frac{\pi}{4}}_0 2tan^2\theta d\theta$

$= \int^{\frac{\pi}{4}}_0 2sec^2\theta - 2 d\theta$

$= [2tan\theta - 2\theta]^{\frac{\pi}{4}}_0$

$= 2 - \frac{\pi}{2} - 0$

$= 2 - \frac{\pi}{2}$

Carrotsticks · Apr 28, 2012

$\\ \int \frac{\sqrt x}{1+x} \\\\$

$\\ x=\tan^2 \theta \\\\ dx=2 \tan \theta \sec^2 \theta d \theta$

$\\ \int \frac{\sqrt x}{1+x} \\\\ = \int \frac{\tan \theta}{1+\tan^2 \theta} \times 2 \tan \theta \sec^2 \theta d \theta \\\\ = 2 \int \frac{\tan^2 \theta \sec^2 \theta}{1+\tan^2 \theta} d \theta \\\\ = 2 \int \frac{\tan^2 \theta \sec^2 \theta}{\sec^2 \theta} d \theta \\\\ = 2 \int \tan^2 \theta d \theta \\\\ = 2 \int \sec^2 \theta - 1 d \theta \\\\ = 2 \left ( \tan \theta - \theta \right ) + C$

And I am too slow!

mirakon · Apr 28, 2012

Carrotsticks said:
$\\ \int \frac{\sqrt x}{1+x} \\\\$

$\\ x=\tan^2 \theta \\\\ dx=2 \tan \theta \sec^2 \theta d \theta$

$\\ \int \frac{\sqrt x}{1+x} \\\\ = \int \frac{\tan \theta}{1+\tan^2 \theta} \times 2 \tan \theta \sec^2 \theta d \theta \\\\ = 2 \int \frac{\tan^2 \theta \sec^2 \theta}{1+\tan^2 \theta} d \theta \\\\ = 2 \int \frac{\tan^2 \theta \sec^2 \theta}{\sec^2 \theta} d \theta \\\\ = 2 \int \tan^2 \theta d \theta \\\\ = 2 \int \sec^2 \theta - 1 d \theta \\\\ = 2 \left ( \tan \theta - \theta \right ) + C$

And I am too slow!

you also didn't read the question properly

it was a definite integral haha

Carrotsticks · Apr 28, 2012

mirakon said:
you also didn't read the question properly it was a definite integral haha

0 marks for me!

Peeik · Apr 28, 2012

Carrotsticks said:
$\\ \int \frac{\sqrt x}{1+x} \\\\$

$\\ x=\tan^2 \theta \\\\ dx=2 \tan \theta \sec^2 \theta d \theta$

$\\ \int \frac{\sqrt x}{1+x} \\\\ = \int \frac{\tan \theta}{1+\tan^2 \theta} \times 2 \tan \theta \sec^2 \theta d \theta \\\\ = 2 \int \frac{\tan^2 \theta \sec^2 \theta}{1+\tan^2 \theta} d \theta \\\\ = 2 \int \frac{\tan^2 \theta \sec^2 \theta}{\sec^2 \theta} d \theta \\\\ = 2 \int \tan^2 \theta d \theta \\\\ = 2 \int \sec^2 \theta - 1 d \theta \\\\ = 2 \left ( \tan \theta - \theta \right ) + C$

And I am too slow!

And you forgot the important dx in your first line. How could you carrot!!

Drongoski · Apr 28, 2012

Just saw the question. Here's my solution without the bothersome substitution.

$\int _0 ^1 \frac {\sqrt x}{1+ x} dx = \int _0 ^1 \frac {\sqrt x}{1+x}d (\sqrt x)^2 = \int _0 ^1 \frac {\sqrt x \times 2 \sqrt x }{1+x}d\sqrt x$

$= \int _0 ^1 \frac {2x+2 - 2}{1+x} d \sqrt x = \int _0 ^1 [2 - 2\frac {1}{1+(\sqrt x)^2}] d \sqrt x = [2\sqrt x -2 tan^{-1}\sqrt x]_0 ^1$

$= 2 - \frac {\pi}{2}$

Notice without substitution, I also didn't need to find new limits.

On reflection, for this question, the substitution method is the easiest. My solution looks tortured, if not downright retarded.

imsleepyz · Apr 28, 2012

How did you guys know it would involve a trig substitution?

Carrotsticks · Apr 28, 2012

imsleepyz said:
How did you guys know it would involve a trig substitution?

Because OP told us to =p

Otherwise Drongoski's method is far more preferable.

nfreidman said:
how do you integate

sqrt (x)/1+x as a definite integral between 1 and zero, and with the substition x=tan^2(theta)

thank you!

mirakon · Apr 28, 2012

imsleepyz said:
How did you guys know it would involve a trig substitution?

the OP states the substitution required

However, even if this wasn't the case you can also deduce it intuitively from practice.

The more questions you do, the easier for you to get used to when particular substitutions are appropriate.

imsleepyz · Apr 28, 2012

mirakon said:
the OP states the substitution required

However, even if this wasn't the case you can also deduce it intuitively from practice.

The more questions you do, the easier for you to get used to when particular substitutions are appropriate.

Yeah forgot that was a given! Anyway I tried to use another method by letting u=sqrt(x) and it worked.

nfreidman · Apr 28, 2012

thank you!! and what about this one:

integrate cube root of(sinx.cos^3x)

Drongoski · Apr 28, 2012

$\int (sinx cos^3x)^{\frac {1}{3}}dx = \int sin ^{\frac {1}{3}}x cos x dx = \frac {3}{4} sin ^{\frac {4}{3}} + C=$

Just saw above question. Didn't know it was done ages ago in a separate thread.

integration (1 Viewer)

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