What is the integral of 7cos(x)(2tan(x)+4sec(x)) ?

The integral of 7cos(x)(2tan(x)+4sec(x)) is 14(-2cos^2(x/2)+4arctan(tan(x/2)))+C

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integral of 7cos(x)(2tan(x)+4sec(x))

\int 7 cos (x) (2 tan (x) + 4 sec (x)) d x

14 (- 2 cos^{2} (\frac{x}{2}) + 4 arctan (tan (\frac{x}{2}))) + C

Solution steps

\int 7 cos (x) (2 tan (x) + 4 sec (x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int cos (x) (2 tan (x) + 4 sec (x)) d x

Apply u-substitution

: \int \frac{2 ( 4 u + 4 ( 1 + u ^{2} ) )}{( 1 + u ^{2} ) ^{2}} d u

= 7 \cdot \int \frac{2 ( 4 u + 4 ( 1 + u ^{2} ) )}{( 1 + u ^{2} ) ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot 2 \cdot \int \frac{4 u + 4 ( 1 + u ^{2} )}{( 1 + u ^{2} ) ^{2}} d u

Expand

\frac{4 u + 4 ( 1 + u ^{2} )}{( 1 + u ^{2} ) ^{2}} : \frac{4 u}{( 1 + u ^{2} ) ^{2}} + \frac{4 ( 1 + u ^{2} )}{( 1 + u ^{2} ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 \cdot 2 (\int \frac{4 u}{( 1 + u ^{2} ) ^{2}} d u + \int \frac{4 ( 1 + u ^{2} )}{( 1 + u ^{2} ) ^{2}} d u)

\int \frac{4 u}{( 1 + u ^{2} ) ^{2}} d u = - \frac{2}{1 + u ^{2}}

\int \frac{4 ( 1 + u ^{2} )}{( 1 + u ^{2} ) ^{2}} d u = 4 arctan (u)

= 7 \cdot 2 (- \frac{2}{1 + u ^{2}} + 4 arctan (u))

Substitute back

u = tan (\frac{x}{2})

= 7 \cdot 2 (- \frac{2}{1 + tan ^{2} ( \frac{x}{2} )} + 4 arctan (tan (\frac{x}{2})))

Simplify

7 \cdot 2 (- \frac{2}{1 + tan ^{2} ( \frac{x}{2} )} + 4 arctan (tan (\frac{x}{2}))) : 14 (- 2 cos^{2} (\frac{x}{2}) + 4 arctan (tan (\frac{x}{2})))

= 14 (- 2 cos^{2} (\frac{x}{2}) + 4 arctan (tan (\frac{x}{2})))

Add a constant to the solution

= 14 (- 2 cos^{2} (\frac{x}{2}) + 4 arctan (tan (\frac{x}{2}))) + C

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What is the integral of 7cos(x)(2tan(x)+4sec(x)) ?
The integral of 7cos(x)(2tan(x)+4sec(x)) is 14(-2cos^2(x/2)+4arctan(tan(x/2)))+C