Math, asked by HridayAg0102, 11 months ago

Integrate
∫ cos(2x). ㏑(1 + tanx)

Answers

Answered by rohitkumargupta
23

HELLO DEAR,

GIVEN:- I = ∫ cos(2x). ㏑(1 + tanx)

Integrate by parts by taking cos2x as second function & log (1 + tanx) as second function

=> log(1 + tanx). sin2x/2 - 1/(1 + tanx).sec²x . sin2x/2 .dx

=> log(1 + tanx).sin2x/2 - tanx/(1 + tanx).dx

=> sinx.cosx.log(1 + tanx) - (1+ tanx)/(1 + tanx).dx + 1/(1 + tanx).dx

=> sinx.cosx.log(1 + tanx) - dx + 1/(1 + tanx).dx

=> sinx.cosx.log(1 + tanx) - x + 1/(1 + tanx).dx

=> sinx.cosx.log(1 + tanx) - x + cosx.dx/(cosx + sinx)

=> sinx.cosx.log(1 + tanx) - x + 1/2(2cosx)/(cosx + sinx).dx

=> sinx.cosx.log(1 + tanx) - x + 1/2{(cosx + sinx) + (cosx - sinx)}/(sinx + cosx).dx

=> sinx.cosx.log(1 + tanx) - x + 1/2dx + 1/2(cosx - sinx)/(sinx + cosx).dx

[as, we know d(sinx + cosx)/dx = cosx - sinx]

=> sinx.cosx.log(1 + tanx) - 1/2x + 1/2log(sinx + cosx) + C

where, C is constant term.

I HOPE IT'S HELP YOU DEAR,

THANKS


tiwaavi: wow. Good answer...
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