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Question #33f91

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Shwetank Mauria

Mar 6, 2017

#int_0^(pi/4)secx^2tanxdx=1/2#

Explanation:

Perhaps you mean #int_0^(pi/4)secx^2tanxdx#

To find this assume #u=tanx#, then #du=sec^2xdx# and

Observe that as #tan0=0# and #tan(pi/4)=1#, the new limits are #0# and #1# and

#int_0^(pi/4)secx^2tanxdx=int_0^1udu#

= #[u^2/2]_0^1#

= #1/2-0=1/2#

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