# Thread: Reduction Integral Formula (tan)

1. ## Reduction Integral Formula (tan)

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

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

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

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

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

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

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

__________________
Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance.
$\displaystyle I_n(x) = \displaystyle\int^x_0 \tan ^n \theta \, \mathrm{d}\theta$

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

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

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

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

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

Bobak

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

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

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

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

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

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

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

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.

4. 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.

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

Bobak

5. Hello,

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

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

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

__________________
Help? Which to integrate and differentiate and when you do these what do you get? Thanks in advance.
If you want to integrate by parts and use the given indication... :

\displaystyle \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