Definite Integrals of Trig Functions

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Question 1 of 2

Find the integral

\int_{\frac{π}{4}}^{\frac{π}{2}} cos x d x

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;\text{cos}x\;dx$

1. $\frac{1}{\sqrt{2}}$ $1/(sqrt2)$
2. $\sqrt{2}$ $sqrt2$
3. $1 - \frac{1}{\sqrt{2}}$ $1-1/(sqrt2)$
4. $1 - \sqrt{2}$ $1-sqrt2$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

First, integrate the trigonometric function

\int_{\frac{π}{4}}^{\frac{π}{2}} cos x d x

$\int_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}\;\text{cos}\;x\;dx$

=

$=$

[sin x]_{\frac{π}{4}}^{\frac{π}{2}}

$\bigg[\text{sin}\;x\bigg]_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}$

Integrate

cos x

$\text(cos)x$

Finally, get the difference of the upper and lower limits substituted to the integral as

x

$x$ .

	$\bigg[\text{sin}\;x\bigg]_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}$

$=$ $=$	$\sin{\color{#9a00c7}{\frac{\pi}{2}}}-\sin{\color{#00880A}{\frac{\pi}{4}}}$	Substitute the limits

$=$	$1-1/(sqrt2)$	Evaluate

$1-1/(sqrt2)$

Question 2 of 2

Find the integral

$\int_{0}^{\frac{\pi}{3}}\;3\;\text{sin}\frac{x}{2}\;dx$

1. $3sqrt3-6$
2. $-3sqrt3+6$
3. $3sqrt3+6$
4. $-3sqrt3-6$

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Integrals of Trigonometric Functions

$int \text(cos)=\text(sin)$

$int \text(sin)=-\text(cos)$

$int \text(sec)^2=\text(tan)$

First, integrate the trigonometric function

$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}\;\text{sin}\;\frac{x}{2}\;dx$	$=$	$\bigg[3\left(-\text{cos}\;\frac{x}{2}\right)\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$	Integrate $3\text(sin) x/2$

	$=$	$\bigg[-6\text{cos}\;\frac{x}{2}\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$	Simplify

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$\bigg[-6\text{cos}\;\frac{x}{2}\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$

$=$	$-6\;\cos{\frac{\color{#9a00c7}{\frac{\pi}{3}}}{2}}-[-6\;\cos{\frac{\color{#00880A}{0}}{2}}]$	Substitute the limits

$=$	$-6 \text(cos)(pi/6)+(6*1)$	Evaluate

$=$	$-6*(sqrt3)/2+6$	$\text(cos) pi/6=(sqrt3)/2$

$=$	$-3sqrt3+6$	Simplify

$-3sqrt3+6$

Definite Integrals of Trig Functions

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1. Question

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Fantastic!

Integrals of Trigonometric Functions

2. Question

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Well Done!

Integrals of Trigonometric Functions

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