Integral of 1cos(x)\frac{1}{\cos{\left(x \right)}}

The calculator will find the integral/antiderivative of 1cos(x)\frac{1}{\cos{\left(x \right)}}, with steps shown.

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Find 1cos(x)dx\int \frac{1}{\cos{\left(x \right)}}\, dx.

Solution

Rewrite the cosine in terms of the sine using the formula cos(x)=sin(x+π2)\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right) and then rewrite the sine using the double angle formula sin(x)=2sin(x2)cos(x2)\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right):

1cos(x)dx=12sin(x2+π4)cos(x2+π4)dx{\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}}}

Multiply the numerator and denominator by sec2(x2+π4)\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right):

12sin(x2+π4)cos(x2+π4)dx=sec2(x2+π4)2tan(x2+π4)dx{\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}}}

Let u=tan(x2+π4)u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}.

Then du=(tan(x2+π4))dx=sec2(x2+π4)2dxdu=\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 (steps can be seen »), and we have that sec2(x2+π4)dx=2du\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du.

Thus,

sec2(x2+π4)2tan(x2+π4)dx=1udu{\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}}}

The integral of 1u\frac{1}{u} is 1udu=ln(u)\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}:

1udu=ln(u){\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}

Recall that u=tan(x2+π4)u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}:

ln(u)=ln(tan(x2+π4))\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}

Therefore,

1cos(x)dx=ln(tan(x2+π4))\int{\frac{1}{\cos{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}

Add the constant of integration:

1cos(x)dx=ln(tan(x2+π4))+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

Answer

1cos(x)dx=ln(tan(x2+π4))+C\int \frac{1}{\cos{\left(x \right)}}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right) + CA