# Find: <img

Let

Now, we know that, sin2x + cos2x = 1

Therefore,

Also, sin2x = 2.sinx.cosx

Therefore,

Now, we know that,

Hence,

OR

Let

Let sin-12x = t, 2x = sint

Differentiating both sides, we get,

2dx = cost dt

Therefore,

Now, Integrating, by parts, taking t as 1st function and cost as the second function.

We know that,

Therefore,

Putting the value of t, we get,

Hence,

I = x.sin-12x – x + C

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