# Reduction Integral Formula (tan)

• May 8th 2008, 10:49 AM
Simplicity
Reduction Integral Formula (tan)
Q: $I_n(x) = \displaystyle\int^x_0 \tan ^x \theta \, \mathrm{d}\theta, \ \ n \ge 0, |x|<\frac{\pi}{2}$

By writing $\tan \theta$ as $\tan ^{n-2} \theta \tan ^2 \theta$, or otherwise, show that

$I_x(x) = \frac{1}{n-1} \tan ^{n-1}x - I_{n-2}(x), \ \ n \ge 2, |x| <\frac{\pi}{2}$

__________________
Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance. (Smile)
• May 8th 2008, 10:59 AM
bobak
Quote:

Originally Posted by Air
Q: $I_n(x) = \displaystyle\int^x_0 \tan ^x \theta \, \mathrm{d}\theta, \ \ n \ge 0, |x|<\frac{\pi}{2}$

By writing $\tan \theta$ as $\tan ^{n-2} \theta \tan ^2 \theta$, or otherwise, show that

$I_x(x) = \frac{1}{n-1} \tan ^{n-1}x - I_{n-2}(x), \ \ n \ge 2, |x| <\frac{\pi}{2}$

__________________
Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance. (Smile)

$I_n(x) = \displaystyle\int^x_0 \tan ^n \theta \, \mathrm{d}\theta$

$I_n(x) = \displaystyle\int^x_0 \tan^{n-2}\theta (sec^2 \theta -1) \theta \, \mathrm{d}\theta$

$I_n(x) = \int^x_0 \tan^{n-2}\theta (sec^2 \theta-1) \, \mathrm{d}\theta$

$I_n(x) = \int^x_0 \tan^{n-2}\theta sec^2 \theta \, \mathrm{d}\theta - \int^x_0 \tan^{n-2}\theta \mathrm{d}\theta$

$I_n(x) = \frac{1}{n-1} \left[ tan^{n-1} \theta \right]^{x}_{0} - I_{n-2}(x)$

$I_n(x) = \frac{1}{n-1} tan^{n-1}x - I_{n-2}(x)$

Bobak

P.S are you sitting the exam on the 18th of june?
• May 8th 2008, 11:03 AM
Simplicity
Quote:

Originally Posted by bobak
$I_n(x) = \displaystyle\int^x_0 \tan ^n \theta \, \mathrm{d}\theta$

$I_n(x) = \displaystyle\int^x_0 \tan^{n-2}\theta (sec^2 \theta -1) \theta \, \mathrm{d}\theta$

$I_n(x) = \int^x_0 \tan^{n-2}\theta (sec^2 \theta-1) \, \mathrm{d}\theta$

$I_n(x) = \int^x_0 \tan^{n-2}\theta sec^2 \theta \, \mathrm{d}\theta - \int^x_0 \tan^{n-2}\theta \mathrm{d}\theta$

$I_n(x) = \frac{1}{n-1} \left[ tan^{n-1} \theta \right]^{x}_{0} - I_{n-2}(x)$

$I_n(x) = \frac{1}{n-1} tan^{n-1}x - I_{n-2}(x)$

Bobak

P.S are you sitting the exam on the 18th of june?

Oh, so you didn't need to use integration by parts?

And, yes, exam on that day. :)
• May 8th 2008, 11:10 AM
bobak
Quote:

Originally Posted by Air
Oh, so you didn't need to use integration by parts?

And, yes, exam on that day. :)

Reduction formula do not always need the use of parts. If you own the heinemann FP2 book look at question 73 on page 126, you don't even need to use any integration method for that.

$\tan^{n-2}\theta sec^2 \theta$ you should spot as a standard form $f^n(x) f'(x)$ in which cases you could use the subistution $u = f(x)$ or just do it straight off.

Bobak
• May 8th 2008, 01:49 PM
Moo
Hello,

Quote:

Originally Posted by Air
Q: $I_n(x) = \displaystyle\int^x_0 \tan ^x \theta \, \mathrm{d}\theta, \ \ n \ge 0, |x|<\frac{\pi}{2}$

By writing $\tan \theta$ as $\tan ^{n-2} \theta \tan ^2 \theta$, or otherwise, show that

$I_x(x) = \frac{1}{n-1} \tan ^{n-1}x - I_{n-2}(x), \ \ n \ge 2, |x| <\frac{\pi}{2}$

__________________
Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance. (Smile)

If you want to integrate by parts and use the given indication... :

\begin{aligned} I_n(x) &=\int_0^x \tan^n \theta d\theta \\
&=\int_0^x \tan^{n-2} \theta \cdot \tan^2 \theta d \theta \\
&=\int_0^x \tan^{n-2} \theta \cdot (\tan^2 \theta +1)-\tan^{n-2} \theta d \theta \\
&=\int_0^x \tan^{n-2} \theta \cdot (\tan^2 \theta +1) d \theta-I_{n-2}(x)
\end{aligned}

Now, integrate by parts the first term, knowing that the derivative of tan(x) is 1+tanē(x), and you got it (Wink)