The integration of the square root of 1+sin2x is given by

∫√(1+sin2x) dx = cosx-sinx+C

where C is an integration constant. In this post, we will learn how to integrate square root of 1+sin2x.

## Integral of Square Root of 1+sin2x

**Question:** Find the integral of 1+sin2x, that is, find

∫√(1+sin2x) dx.

**Answer:**

We will use the following two trigonometric formulas:

- sin
^{2}x+cos^{2}x = 1 - sin2x= 2sinxcosx

By the above two formulas, we have that

1+sin2x = sin^{2}x+cos^{2}x+2sinxcosx = (sinx+cosx)^{2}

Taking square root of both sides, we obtain that

√(1+sin2x) = sinx+cosx

Integrating both sides, we get that

∫√(1+sin2x) dx = ∫(sinx+cosx) dx + C, C is an integral constant.

= ∫sinx dx + ∫cosx dx + C

= -cosx+sinx+C.

So the integration of square root of 1+sin2x is equal to -cosx+sinx+C where C is an integration constant.

## Integration of root 1+sin2x from 0 to pi/2

**Question:** Find the integral of root 1+sin2x from 0 to π/2.

*Answer:*

From above, we have ∫√(1+sin2x) dx = cosx-sinx+C. Thus, the definite integral of root 1+sin2x from 0 to π/2 will be equal to

$\int_0^{\pi/2} \sqrt{1+\sin 2x} dx$

= $[\cos x -\sin x]_0^{\pi/2}$

= (cos π/2 – sin π/2) – (cos0 – sin0)

= (0-1) – (1-0)

= -1-1

= -2

So the integration of root 1+sin2x from 0 to pi/2 is equal to -2.

**ALSO READ:**

**Integration of log(sinx) from 0 to pi/2**

**Integration of root(x)+1/root(x)**

## FAQs

**Q1: What is the integration of root 1+sin2x?**

**Answer: **The integration of root 1+sin2x is given as follows: ∫√(1+sin2x) dx = cosx-sinx+C, where C is an integral constant.