Question

Evaluate the integral\(\int {\frac{{{\rm{1 - tan\theta }}}}{{{\rm{1 + tan\theta }}}}} {\rm{d\theta }}\).

Short answer

The integral value of the given equation is\(\int {\frac{{{\rm{1 - tan\theta }}}}{{{\rm{1 + tan\theta }}}}} {\rm{d\theta = ln|sin\theta + cos\theta | + C}}\).

Step-by-step solution

Expand the equation.

Given: \(\int {\frac{{{\rm{1 - tan\theta }}}}{{{\rm{1 + tan\theta }}}}} {\rm{d\theta }}\)

Known value\({\rm{tan\theta = }}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}\)

\({\rm{ = }}\int {\frac{{{\rm{1 - }}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}}}{{{\rm{1 + }}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}}}} {\rm{d\theta }}\)

Evaluate the equation.

To get\({\rm{cos\theta }}\), multiply both the numerator and the denominator by\({\rm{cos\theta }}\).

\(\int {\frac{{{\rm{cos\theta - sin\theta }}}}{{{\rm{cos\theta + sin\theta }}}}} {\rm{d\theta }}\)

Substitute \({\rm{u = sin\theta + cos\theta }}\)and \({\rm{du = (cos\theta - sin\theta )d\theta }}\)

\({\rm{ = }}\int {\frac{{{\rm{du}}}}{{\rm{u}}}} {\rm{ = ln|u| + C}}\)

Return to the original position\({\rm{u = sin\theta + cos\theta }}\). To get

\( = \ln |\sin \theta + \cos \theta | + C\)

Therefore, the integral value of the given equation is\(\int {\frac{{{\rm{1 - tan\theta }}}}{{{\rm{1 + tan\theta }}}}} {\rm{d\theta = ln|sin\theta + cos\theta | + C}}\).