Question

If \(\rm I_n = \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^n \theta \ d\theta \), then I₈ + I₆ equals:
1. \(\dfrac14\)
2. \(\dfrac15\)
3. \(\dfrac16\)
4. \(\dfrac17\)

Answer 1

Correct Answer - Option 4 : \(\dfrac17\)

Concept:

Integration by Parts:

• ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ f'(x) [∫ g(x) dx] dx.

 

Integration by substitution:

• If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt.

 

Definite Integral:

• If ∫ f(x)dx = g(x) + C, then \(\rm \displaystyle \int_a^b f(x)dx = [ g(x)]_a^b\) = g(b) - g(a).

 

Derivatives of Trigonometric Functions:

• \(\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc\)

 

Trigonometric identities:

• sin² θ + cos² θ = 1.
• tan2 θ + 1 = sec2 θ.

 

Calculation:

Let us first consider \(\rm \displaystyle\int \tan^8 x \ dx\).

= \(\rm \displaystyle\int \tan^6 x \tan^2 x \ dx\)

= \(\rm \displaystyle\int \tan^6 x (\sec^2 x-1) \ dx\)

= \(\rm \displaystyle\int \tan^6 x \sec^2 x\ dx-\int \tan^6 x \ dx\)

Now, let's consider \(\rm \displaystyle\int \tan^6 x \sec^2 x \ dx\).

Substitute tan x = u ⇒ sec² x dx = du.

∴ \(\rm \displaystyle\int \tan^6 x \sec^2 x \ dx=\int u^6\ du=\dfrac{u^7}{7}+C=\dfrac{\tan^7x}{7}+C\)

And, \(\rm \displaystyle\int \tan^8 x \ dx=\dfrac{\tan^7x}{7}+C-\int \tan^6 x \ dx\)

⇒ \(\rm \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^8 x \ dx=\left [\dfrac{\tan^7x}{7} \right ]_0^{\tfrac{\pi}{4}}-\int_0^{\tfrac{\pi}{4}} \tan^6 x \ dx\)

⇒ \(\rm \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^8 x \ dx+\int_0^{\tfrac{\pi}{4}} \tan^6 x \ dx=\left [\dfrac{\tan^7 \tfrac{\pi}{4}}{7} - \dfrac{\tan^7 0}{7}\right ]\)

Using \(\rm \tan\left( \dfrac{\pi}{4}\right) = 1\) and tan 0 = 0, we get:

⇒ \(\rm \displaystyle I_8+I_6=\dfrac{1}{7}\).