If I 1 - 0 - 4 sin 2 x dx and I 2 - 0 - 4 cos 2 x dx- then

The correct option is B I _ 1 For x∈( 0, π nbsp; 4 nbsp;) cos^ 2 x gt; sin^ 2 x ⇒∫_ 0 ^ π nbsp;/ 4 cos^ 2 x dx gt; ∫_ 0 ^ π ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Using Lenz's Law

If I 1 =∫ 0...

Question

If $I_{1} = \int_{0}^{π / 4} {sin}^{2} x d x$ and $I_{2} = \int_{0}^{π / 4} {cos}^{2} x d x$ , then

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

I_{1} > I_{2}

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

I_{2} = I_{1} + \frac{π}{4}

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

Open in App

Solution

Suggest Corrections

Similar questions

I_{1} = \int_{0}^{1} \frac{{tan}^{- 1} x d x}{x}

I_{2} = \int_{0}^{\frac{π}{2}} \frac{x d x}{sin x}

\frac{I_{1}}{I_{2}}

I_{1} = \int_{0}^{π / 2} \frac{x}{sin x} d x

I_{2} = \int_{0}^{π / 2} \frac{{tan}^{- 1} x}{x} d x,

\frac{I_{1}}{I_{2}} =

I_{1} = \int_{0}^{3 π} f ({cos}^{2} x) d x

I_{2} = \int_{0}^{π} f ({cos}^{2} x) d x

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?

I_{1} = \int_{0}^{\frac{π}{2}} f (sin 2 x) sin x d x

I_{2} = \int_{0}^{\frac{π}{2}} f (cos 2 x) cos x d x

\frac{I_{1}}{I_{2}}

Join BYJU'S Learning Program