NIMCET 2013 Question Paper with solutions

© examsiri.com

Question : 6 of 120

Marks: +1, -0

If

I_n

=

∫↖{π/4}↙{0} tan^n

θ dθ , then

I_8+I_6

equals :

1/4
1/5
1/6
1/7

Solution:

Given

I_n=∫↖{π/4}\tan^nθdθ

⇒I_{n}=∫↙{0}↖{π/4}\tan^{n-2}θ\tan^{2}θdθ

⇒I_{n}=∫↙{0}↖{π/4}\tan^{n-2}θ(\sec^{2}θ-1)dθ

[∵

\tan^2θ=(\sec^2θ-1)

]

⇒I_{n}=∫↙{0}↖{π/4}\tan^{n-2}θ\sec^{2}θ-∫↙{0}↖{π/4}\tan^{n-2}θdθ

⇒I_{n}+I_{n-2}=∫↙{0}↖{1}t^{n-2}dt=[{t^{n-1}}/{n-1}]_{0}^{1}

[∵ let

\tanθ=t]

⇒I_{n}+I_{n-2}={1}/{n-1}

Now, putting n = 8,we get

I_8+I_{8-2}=1/{8-1}

⇒I_8+I_6=1/7

© examsiri.com

Go to Question:

Prev Question

Next Question