The value of -0-2sin x-cos x2-1-sin 2 x d x is

The correct option is C 2I = ∫_0^π/2 (sin x+cos x)^2/√(1+sin 2x)dx = ∫_0^π/2(sin x+cos x)^2/√(cos^2x+sin^2x+2 sin x cos x)dx =∫_0^π/2(sin x+cos x)dx=[ cos x+si ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

The value of $\int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x i s$

0.0

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

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

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

Open in App

Solution

The correct option is C 2
$I = \int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x = \int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{c o s^{2} x + s i n^{2} x + 2 s i n x c o s x}} d x$
$= \int_{0}^{\frac{π}{2}} (s i n x + c o s x) d x = [- c o s x + s i n x]_{0}^{\frac{π}{2}} = 2$

Suggest Corrections

Similar questions

\int_{0}^{\frac{π}{2}} \frac{(sin x + cos x)^{2}}{\sqrt{1 + sin 2 x}} d x

Q. The value of

I = \int_{0}^{π / 2} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x

Q. Evaluate:

\int_{0}^{π / 2} \frac{{(sin x + cos x)}^{2}}{\sqrt{1 + sin 2 x}} d x

Evaluate

\int_{0}^{π / 2} \frac{sin x + cos x}{3 + sin 2 x} d x

\int_{0}^{\frac{π}{2}} \frac{s i n x c o s x}{1 + s i n^{4} x} d x =

Join BYJU'S Learning Program

The value of ∫π20 (sin x+cos x)2√1+sin 2xdx is

The correct option is C 2I=∫π20 (sin x+cos x)2√1+sin 2xdx=∫π20(sin x+cos x)2√cos2x+sin2x+2 sin x cos xdx =∫π20(sin x+cos x)dx=[−cos x+sin x]π20=2

The value of $\int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x i s$