Evaluate the integrals. \int_{0}^{\pi/12}6\tan 3tdt;\ \int \sin^6xdx

Question:

Evaluate the integrals.

{eq}\int_{0}^{\pi/12}6\tan 3tdt;\ \int \sin^6xdx {/eq}

Integrals:

In order to evaluate the given integrals, we will apply different methods of the integral calculus. In the first part, we will apply the basic rules of the integral calculus, whereas in the second part, we will make use of the integral reduction formula, which is given as: {eq}\begin{align*} \ **** \int \sin^n x \ dx = -\frac{1}{n} \sin ^{(n-1)} x \cos x + \frac{(n-1)}{n} \int \sin^{(n-2)} x \ dx \end{align*} {/eq}.

Answer and Explanation:


Given {eq}\int_{0}^{\pi/12}6\tan 3t \ dt {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = 6\int_{0}^{\pi/12} \frac{\sin 3t}{\cos 3t} \ dt \end{align*} {/eq}

Multiplying and dividing the integral by 3 and "-" sign, we get:

{eq}\begin{align*} \ & = -\frac{6}{3}\int_{0}^{\pi/12} \frac{-3\sin 3t}{\cos 3t} \ dt \\ \\ \ & = -2 \left[\ln |\cos 3t| \right]_{0}^{\pi/12} \ \ \ \ \ \ \ \ \ \ \left[ \int \frac{f'(x)}{f(x)} \ dx = \ln (f (x)) + c \right] \\ \\ \ & = -2 \left[\ln |\cos \frac{3\pi}{12}| - \ln |\cos 0| \right] \\ \\ \ & = -2 \left[\ln |\frac{1}{\sqrt2}| - \ln |1| \right] \\ \\ \ & = -2 \left[ -0.347- 0 \right] \\ \\ \ & = 0.694 \end{align*} {/eq}


Given {eq}\int \sin^6x \ dx {/eq}

Applying integral reduction formula, we get:

{eq}\begin{align*} \ & = -\frac{1}{6} \sin ^5 x \cos x + \frac{5}{6} \int \sin^4 x \ dx \end{align*} {/eq}

Again applying integral reduction formula, we get:

{eq}\begin{align*} \ & = -\frac{1}{6} \sin ^5 x \cos x + \frac{5}{6} \left[ -\frac{1}{4} \sin^3 x \cos x + \frac{3}{4} \int \sin^2 x \ dx \right] \\ \\ \ & = -\frac{1}{6} \sin ^5 x \cos x -\frac{5}{24} \sin^3 x \cos x + \frac{5}{8} \int \sin^2 x \ dx \\ \\ \ & = -\frac{1}{6} \sin ^5 x \cos x -\frac{5}{24} \sin^3 x \cos x + \frac{5}{8} \int \frac{1-\cos 2x}{2} \ dx \ \ \ \ \ \ \ \ \ \ \ \ \left[ \sin^2 x = \frac{1-\cos 2x}{2}\right] \\ \\ \ & = -\frac{1}{6} \sin ^5 x \cos x -\frac{5}{24} \sin^3 x \cos x + \frac{5}{16} \int (1-\cos 2x) \ dx \\ \\ \ & = -\frac{1}{6} \sin ^5 x \cos x -\frac{5}{24} \sin^3 x \cos x + \frac{5}{16} \left(x-\frac{\sin 2x}{2} \right) + c \end{align*} {/eq}


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Basic Calculus: Rules & Formulas

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