Evaluate the integral \int_{\pi/4}^{\pi/3} \tan x \, dx

Question:

Evaluate the integral {eq}\int_{\pi/4}^{\pi/3} \tan x \, dx {/eq}

U-Substitution:

When working with more complicated integrals, some form of substitution may be necessary. With composite functions and many of the trig functions, {eq}u {/eq}-substitution can be helpful.

{eq}\int f\left ( g\left ( x \right ) \right ){g}'\left ( x \right )dx=\int f\left ( u \right )du {/eq}

when {eq}u=g\left ( x \right ) {/eq}.

For definite integrals:

{eq}\int_{a}^{b} f\left ( g\left ( x \right ) \right ){g}'\left ( x \right )dx=\int_{g(a)}^{g(b)} f\left ( u \right )du {/eq}

when {eq}u=g\left ( x \right ) {/eq}.

Answer and Explanation:

To evaluate the integral {eq}\int_{\pi/4}^{\pi/3} \tan x \, dx {/eq}, we will use {eq}u {/eq}-substitution.

{eq}\displaystyle{\int_{\pi/4}^{\pi/3} \tan x \, dx}=\displaystyle{\int_{\pi/4}^{\pi/3}\frac{\sin x}{\cos x}\, dx} {/eq}


Let

{eq}u=\cos x\\ du=-\sin x dx {/eq}

Write the integral in terms of {eq}u {/eq} including the limits of integration. When {eq}x=\frac{\pi}{4} {/eq}, {eq}u=\frac{\sqrt{2}}{2} {/eq}. When {eq}x=\frac{\pi}{3} {/eq}, {eq}u=\frac{1}{2} {/eq}.


{eq}\begin{align*} \displaystyle{-\int_{\frac{\sqrt{2}}{2}}^{\frac{1}{2}}\frac{1}{u}du}&=\displaystyle{-\left ( \ln \left |u \right | \right )\bigg|_{\frac{\sqrt{2}}{2}}^{\frac{1}{2}}}\\ &=-\left ( \ln \frac{1}{2}-\ln \frac{\sqrt{2}}{2} \right )\\ &=\ln \frac{\sqrt{2}}{2}- \ln \frac{1}{2}\\ &=\ln \sqrt{2}\\ &=\boxed{\frac{1}{2}\ln 2} \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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