Question

integration root of 1-cos2x dx

Answer 1

i hope this will help u

Answer 2

Answer:

The correct answer is $& \int \sqrt{1-\cos 2 x} d x$ $=&-\sqrt{2} \cos x+c$.

Step-by-step explanation:

Integration:

Integration exists as a process of adding or summing up the components to find the total. It exists a reverse method of differentiation, where we facilitate the operations.

Given:

$& \int \sqrt{1-\cos 2 x} d x$

We have to prove that

$& \int \sqrt{1-\cos 2 x} d x$

Step 1

Let,

$& \int \sqrt{1-\cos 2 x} d x......................(1)$

$& \cos 2 x=1-2 \sin ^{2} x$

therefore $& 1-\cos 2 x=2 \sin ^{2} x$

Step 2

Substituting the values in $(1)$, we get

$=& \int \sqrt{2} \sin x d x$

$=& \sqrt{2} \int \sin x d x$

When $& \int \sin x d x =-Cos x+C$

then,

$=&-\sqrt{2} \cos x+c$

Therefore, we get

$& \int \sqrt{1-\cos 2 x} d x$ $=&-\sqrt{2} \cos x+c$.

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