integrate 1/1-tanx

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integrate

1/1-tanx

Posted by Vishaal Velu 8 years, 3 months ago

CBSE > Class 12 > Mathematics

1 answers

Naveen Sharma 8 years, 3 months ago

Ans. $\int {1\over 1-tan x}dx$

$=> \int {1\over 1-{sinx\over cosx}}dx$

$=> \int {cos x\over cos x - sinx}dx$

Multiply and Divide By 2, We Get

$=> {1\over2}\int {2cos x\over cos x-sinx}dx$

$=> {1\over2}\int {2cos x+sin x - sin x \over cos x-sinx}dx$

$=> {1\over2}\int {(cos x+sin x)+(cos x - sin x )\over cos x-sinx}dx$

$=> {1\over2}\int {(cos x+sin x)\over cosx-sinx}dx+{1\over2}\int{ (cos x - sin x )\over cos x-sinx}dx$

$=> {1\over2}\int {(cos x+sin x)\over cosx-sinx}dx+{1\over2}\int dx$

Put (cos x - sin x) = t, then

(-sin x - cos x)dx = dt or -(cos x +sin x) dx = dt

$=> {1\over2}\int {-dt\over t }+{1\over2}\int dx$

$=> -{1\over 2} log t + {1\over 2} x + C$

$=> -{1\over 2} log (cos x-sin x) + {1\over 2} x + C$

1Thank You

ANSWER

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