Evaluate the indefinite integrals. \int 4tan^{3}x\,dx \int \frac{\tan^{2}x}{\csc x}\,dx

Question:

Evaluate the indefinite integrals.

{eq}\displaystyle \int 4tan^{3}x\,dx {/eq}

{eq}\displaystyle \int \frac{\tan^{2}x}{\csc x}\,dx {/eq}

Indefinite Integral:

Integral which gives the result with an arbitrary constant is called Indefinite Integral.

In the first given integral, we will use U-Substititution method and the second one will be solved by normal way.

Answer and Explanation:

{eq}\begin{align*} \int 4 \tan^3 (x) \, dx &= 4 \int \tan (x) \tan^2 (x) \, dx \\ &= 4 \int \tan (x) \left( \sec^2 (x) - 1 \right) \, dx \hspace{1 cm} \left[ \because \sec^2 (x) - \tan^2 (x) = 1 \right] \\ &= 4 \int \tan (x) \sec^2 (x) \, dx - 4 \int \tan (x) \, dx \end{align*} {/eq}


Now:

Let us assume that:

{eq}\tan (x) = u \\ \Rightarrow \sec^2 (x) \, dx = du {/eq}


Substituting {eq}u {/eq} into the first integration and rewriting the given integral, we have:

{eq}\begin{align*} \int 4 \tan^3 (x) \, dx &= 4 \int u \, du - 4 \int \tan (x) \, dx \\ &= 2u^2 - 4 \ln \mid \sec (x) \mid + C \hspace{1 cm} \left[ \because \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \, and \, \int \tan (x) \, dx = \ln \mid \sec (x) \mid + C \, \text{ Where C is an integral constant} \right] \end{align*} {/eq}


Undo substitution, we have:

{eq}\int \tan^3 (x) \, dx = 2 \tan^2 (x) - 4 \ln \mid \sec (x) \mid + C {/eq}



{eq}\begin{align*} \int \frac{\tan^2 (x)}{\csc (x)} \, dx &= \int \tan^2 (x) \sin (x) \, dx \hspace{1 cm} \left[ \because \csc (x) = \frac{1}{\sin (x)} \right] \\ &= \int \left( \sec^2 (x) - 1 \right) \sin (x) \, dx \hspace{1 cm} \left[ \because \sec^2 (x) - \tan^2 (x) = 1 \right] \\ &= \int \sec^2 (x) \sin (x) \, dx - \int \sin (x) \, dx \\ &= \int ( \frac{\sin (x)}{\cos (x)} )( \frac{1}{\cos (x)}) \, dx - \int \sin (x) \, dx \hspace{1 cm} \left[ \because \sec^2 (x) = \frac{1}{\cos^2 (x)} \right] \\ &= \int \sec (x) \tan (x) \, dx - \int \sin (x) \, dx \hspace{1 cm} \left[ \because \sec (x) = \frac{1}{\cos (x)} \, and \, \tan (x) = \frac{\sin (x)}{\cos (x)} \right] \\ &= \sec (x) + \cos (x) + C \hspace{1 cm} \left[ \because \int \sin (x) \, dx = - \cos (x) + C \, and \, \int \sec (x) \tan (x) \, dx = \sec (x) + C \, \text{ Where C is an integral constant} \right] \end{align*} {/eq}


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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