What is the integral of 1/((cos^2(x)(1-tan(x))^2)) ?

The integral of 1/((cos^2(x)(1-tan(x))^2)) is -1/(tan(x)-1)+C

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integral of 1/((cos^2(x)(1-tan(x))^2))

integral

\frac{1}{( cos ^{2} ( x ) ( 1 - tan ( x ) ) ^{2} )}

- \frac{1}{tan ( x ) - 1} + C

Solution steps

\int \frac{1}{cos ^{2} ( x ) ( 1 - tan ( x ) ) ^{2}} d x

Rewrite using trig identities

= \int \frac{1}{( cos ( x ) - sin ( x ) ) ^{2}} d x

(cos (x) - sin (x))^{2} = cos^{2} (x) - 2 cos (x) sin (x) + sin^{2} (x)

= \int \frac{1}{cos ^{2} ( x ) - 2 cos ( x ) sin ( x ) + sin ^{2} ( x )} d x

Factor

\frac{1}{cos ^{2} ( x ) - 2 cos ( x ) sin ( x ) + sin ^{2} ( x )}

= \int \frac{1}{( sin ( x ) - cos ( x ) ) ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{( u - 1 ) ^{2}} d u

= \int \frac{1}{( u - 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

= - \frac{1}{tan ( x ) - 1}

Add a constant to the solution

= - \frac{1}{tan ( x ) - 1} + C

\frac{7 x + 2}{1 - 25 x ^{2}}

a^{2} (e^{- a})^{2} x^{4}

\frac{ln ( 4 x )}{x ^{4}}

\frac{1}{( 5 + x ^{2} )}

e^{2} x - 1

What is the integral of 1/((cos^2(x)(1-tan(x))^2)) ?
The integral of 1/((cos^2(x)(1-tan(x))^2)) is -1/(tan(x)-1)+C