Evaluate- -1-cos x sin x-2cos x dx

The correct option is A log ( tan x+2 ) Let nbsp;I=∫ nbsp; 1 /cos nbsp;x( sin nbsp;x+2cos nbsp;x ) nbsp; dx=∫ 1 / cos^ 2 x ( tan x +2 ) ...

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Standard X

Mathematics

Trigonometric Identities

Evaluate: ∫...

Question

Evaluate: $\int \frac{1}{cos x (sin x + 2 cos x)} d x$

log (tan x + 2)

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log (tan x - 2)

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log (sec x + 2)

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- log (tan x + 2)

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Solution

The correct option is A $log (tan x + 2)$
Let $I = \int \frac{1}{cos x (sin x + 2 cos x)} d x = \int \frac{1}{c o s^{2} x (tan x + 2)} d x$
Put $tan x + 2 = t \Rightarrow {sec}^{2} x d x = d t$
Therefore
$I = \int \frac{d t}{t} = log t = log (tan x + 2)$

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Evaluate: ∫1cosx(sinx+2cosx)dx

The correct option is A log(tanx+2)Let I=∫1cosx(sinx+2cosx)dx=∫1cos2x(tanx+2)dxPut tanx+2=t⇒sec2xdx=dtThereforeI=∫dtt=logt=log(tanx+2)

Evaluate: $\int \frac{1}{cos x (sin x + 2 cos x)} d x$

The correct option is A $log (tan x + 2)$
Let $I = \int \frac{1}{cos x (sin x + 2 cos x)} d x = \int \frac{1}{c o s^{2} x (tan x + 2)} d x$
Put $tan x + 2 = t \Rightarrow {sec}^{2} x d x = d t$
Therefore
$I = \int \frac{d t}{t} = log t = log (tan x + 2)$