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Q43 of 261 Page 811

Evaluate the following integrals

Let

Let

Let x=tan t

⇒ dx=sec²tdt.

Also when x=0, t=0 and when x=1, .

Hence,

Using , we get

Let

Let 1+x²=t ⇒ 2xdx=dt.

When x=0, t=1 and when x=1, t=2.

Substituting t=1+x²

⇒ 2xdx=dt.

When t=1, x=0 and when t=2, x=1.

Hence,

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