The calculator will find the integral/antiderivative of
1 cos ( x ) \frac{1}{\cos{\left(x \right)}} c o s ( x ) 1 , with steps shown.
Related calculator:
Definite and Improper Integral Calculator
Solution Rewrite the cosine in terms of the sine using the formula cos ( x ) = sin ( x + π 2 ) \cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right) cos ( x ) = sin ( x + 2 π ) and then rewrite the sine using the double angle formula sin ( x ) = 2 sin ( x 2 ) cos ( x 2 ) \sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) sin ( x ) = 2 sin ( 2 x ) cos ( 2 x ) :
∫ 1 cos ( x ) d x = ∫ 1 2 sin ( x 2 + π 4 ) cos ( x 2 + π 4 ) d x {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} ∫ c o s ( x ) 1 d x = ∫ 2 s i n ( 2 x + 4 π ) c o s ( 2 x + 4 π ) 1 d x
Multiply the numerator and denominator by sec 2 ( x 2 + π 4 ) \sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right) sec 2 ( 2 x + 4 π ) :
∫ 1 2 sin ( x 2 + π 4 ) cos ( x 2 + π 4 ) d x = ∫ sec 2 ( x 2 + π 4 ) 2 tan ( x 2 + π 4 ) d x {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} ∫ 2 s i n ( 2 x + 4 π ) c o s ( 2 x + 4 π ) 1 d x = ∫ 2 t a n ( 2 x + 4 π ) s e c 2 ( 2 x + 4 π ) d x
Let u = tan ( x 2 + π 4 ) u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)} u = tan ( 2 x + 4 π ) .
Then d u = ( tan ( x 2 + π 4 ) ) ′ d x = sec 2 ( x 2 + π 4 ) 2 d x du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx d u = ( tan ( 2 x + 4 π ) ) ′ d x = 2 s e c 2 ( 2 x + 4 π ) d x (steps can be seen » ), and we have that sec 2 ( x 2 + π 4 ) d x = 2 d u \sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du sec 2 ( 2 x + 4 π ) d x = 2 d u .
Thus,
∫ sec 2 ( x 2 + π 4 ) 2 tan ( x 2 + π 4 ) d x = ∫ 1 u d u {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}} ∫ 2 t a n ( 2 x + 4 π ) s e c 2 ( 2 x + 4 π ) d x = ∫ u 1 d u
The integral of 1 u \frac{1}{u} u 1 is ∫ 1 u d u = ln ( ∣ u ∣ ) \int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)} ∫ u 1 d u = ln ( ∣ u ∣ ) :
∫ 1 u d u = ln ( ∣ u ∣ ) {\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}} ∫ u 1 d u = l n ( ∣ u ∣ )
Recall that u = tan ( x 2 + π 4 ) u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)} u = tan ( 2 x + 4 π ) :
ln ( ∣ u ∣ ) = ln ( ∣ tan ( x 2 + π 4 ) ∣ ) \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)} ln ( ∣ u ∣ ) = ln ( ∣ ∣ t a n ( 2 x + 4 π ) ∣ ∣ )
Therefore,
∫ 1 cos ( x ) d x = ln ( ∣ tan ( x 2 + π 4 ) ∣ ) \int{\frac{1}{\cos{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} ∫ cos ( x ) 1 d x = ln ( ∣ ∣ tan ( 2 x + 4 π ) ∣ ∣ )
Add the constant of integration:
∫ 1 cos ( x ) d x = ln ( ∣ tan ( x 2 + π 4 ) ∣ ) + C \int{\frac{1}{\cos{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C ∫ cos ( x ) 1 d x = ln ( ∣ ∣ tan ( 2 x + 4 π ) ∣ ∣ ) + C
Answer ∫ 1 cos ( x ) d x = ln ( ∣ tan ( x 2 + π 4 ) ∣ ) + C \int \frac{1}{\cos{\left(x \right)}}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right) + C ∫ c o s ( x ) 1 d x = ln ( ∣ ∣ tan ( 2 x + 4 π ) ∣ ∣ ) + C A