Math Doubts

Find $\displaystyle \int{\dfrac{1+\tan^2{x}}{1+\cot^2{x}}}dx$

It is a trigonometric function which contains tan and cot functions in fraction form in terms of $x$. We have to find the indefinite integration of the trigonometric function with respect to $x$ in this indefinite integral problem.

Change the denominator same as numerator

The numerator and denominator of the fraction is in similar form but they contain different functions. Actually, they are reciprocal functions. So, change any trigonometric function in its reciprocal form by reciprocal identity of trigonometry.

$= \,\,\,$ $\displaystyle \int{\dfrac{1+\tan^2{x}}{1+\dfrac{1}{\tan^2{x}}}}dx$

Now, simplify the fraction to perform the integration of the trigonometric function in the next few steps.

$= \,\,\,$ $\displaystyle \int{\dfrac{1+\tan^2{x}}{\dfrac{\tan^2{x}+1}{\tan^2{x}}}}dx$

$= \,\,\,$ $\displaystyle \int{(1+\tan^2{x}) \times \dfrac{\tan^2{x}}{\tan^2{x}+1}}dx$

$= \,\,\,$ $\displaystyle \int{\tan^2{x} \times \dfrac{1+\tan^2{x}}{\tan^2{x}+1}}dx$

$= \,\,\,$ $\displaystyle \int{\tan^2{x} \times \dfrac{1+\tan^2{x}}{1+\tan^2{x}}}dx$

$= \,\,\,$ $\require{cancel} \displaystyle \int{\tan^2{x} \times \dfrac{\cancel{1+\tan^2{x}}}{\cancel{1+\tan^2{x}}}}dx$

$= \,\,\,$ $\displaystyle \int{\tan^2{x} \times 1}dx$

$= \,\,\,$ $\displaystyle \int{\tan^2{x}}dx$

Change denominator same as numerator

There is no integral rule to find the integration of square of tan function but it can be done by transforming into square of secant function by using Pythagorean identity of secant and tangent functions.

$= \,\,\,$ $\displaystyle \int{(\sec^2{x}-1)}dx$

Find the indefinite integral of each function

According to difference rule of integration, the integration of difference of two functions is equal to difference of their integrals.

$= \,\,\,$ $\displaystyle \int{\sec^2{x}}dx-\int{1}dx$

Find the integration by using integral of square of secant function and integral of one formulas.

$= \,\,\,$ $\tan{x}-x+c$



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