If fx is an integrable function in -6- -3 and I1- - -6- -3 2x f 2sin 2x dx and I2- - -6- -3co 2x f 2sin 2x dx- then-

The correct option is C I_1= I_2 Let I _ 1 =∫ _π/6^π/3^ 2 xf( 2sin2x ) dx #160;Using property ∫ _ a ^ b f( x ) dx =∫ _ a ^ b f( a+b x ) dx ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration as Antiderivative

If fx is an...

Question

If $f (x)$ is an integrable function in $(\frac{π}{6}, \frac{π}{3})$ and $I_{1} = \int_{π / 6}^{π / 3} {sec}^{2} x f (2 sin 2 x) d x$ and $I_{2} = \int_{π / 6}^{π / 3} c o {sec}^{2} x f (2 sin 2 x) d x,$ then:

I_{1} = 2 I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 I_{1} = I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} = I_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $I_{1} = I_{2}$
Let $I_{1} = \int_{π / 6}^{π / 3} {sec}^{2} x f (2 sin 2 x) d x$
Using property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I_{1} = \int_{π / 6}^{π / 3} {sec}^{2} (\frac{π}{2} - x) f (2 sin 2 (\frac{π}{2} - x)) d x$

$= \int_{π / 6}^{π / 3} {csc}^{2} x f (2 sin 2 x) d x = I_{2}$

Suggest Corrections

Similar questions

If $I_{1} = \int_{0}^{\frac{π}{2}} x s i n x d x & I_{2} = \int_{0}^{\frac{π}{2}} x c o s x d x$ , then which of the following is true?

Q. Let

I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x

, then

Q. If

I_{1} = \int_{0}^{π} x sin x e^{cos 4 x} d x & I_{2} = \int_{0}^{\frac{π}{2}} cos x e^{cos 4 x} d x

, then the value of

[\frac{I_{1}}{I_{2}}]

is (where [.] denotes the greatest integer function)

I_{1} = \int_{0}^{\frac{π}{2}} l n (s i n x) d x, I_{2} = \int_{- \frac{π}{4}}^{\frac{π}{4}} ln (sin x + cos x) d x

.Then

Let $I_{1} = {(\frac{π}{4})}^{2} + \sqrt{2}, I_{2} = {(t a n^{- 1} (\frac{1}{e}))}^{2} + \frac{2 e}{\sqrt{e^{2} + 1}}, I_{3} = (t a n^{- 1} e)^{2} + \frac{2}{\sqrt{e^{2} + 1}}$ , then which of the following is true?

Join BYJU'S Learning Program

If f(x) is an integrable function in (π6,π3) and I1=∫π/3π/6sec2xf(2sin2x)dx and I2=∫π/3π/6cosec2xf(2sin2x)dx, then:

The correct option is C I1=I2Let I1=∫π/3π/6sec2xf(2sin2x)dx Using property ∫baf(x)dx=∫baf(a+b−x)dxI1=∫π/3π/6sec2(π2−x)f(2sin2(π2−x))dx=∫π/3π/6csc2xf(2sin2x)dx=I2

If $f (x)$ is an integrable function in $(\frac{π}{6}, \frac{π}{3})$ and $I_{1} = \int_{π / 6}^{π / 3} {sec}^{2} x f (2 sin 2 x) d x$ and $I_{2} = \int_{π / 6}^{π / 3} c o {sec}^{2} x f (2 sin 2 x) d x,$ then: