Evaluate in fnite terms.\int _a^b cos^4(x)sin^3(x)dx

Question:

Evaluate in finite terms.

{eq}\int _a^b cos^4(x)sin^3(x)dx {/eq}

{eq}\int _a^b \frac{x}{x^2+4x+1}dx {/eq}

{eq}\int^b_acos^6(x)sin^4(x)dx {/eq}

Determination of Anti Derivative of any Function:

Suppose we have two functions {eq}f\left( x \right) {/eq} and {eq}g\left( x \right) {/eq} of one variable, such that {eq}g'\left( x \right) = f\left( x \right) {/eq}

Then we can say that {eq}g\left( x \right) {/eq} is the Antiderivative of the function {eq}f\left( x \right) {/eq}

Hence we can say that: {eq}g\left( x \right) = \int {f\left( x \right)dx} + c {/eq}, where c is the integrating constant.

Again we know that: {eq}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + c {/eq}

Answer and Explanation:

Solution (1):

Here, in this case, the given integral can be solved as:

{eq}\eqalign{ I& = \int_a^b c o{s^4}(x)si{n^3}(x)dx \cr & = \int_a^b {{{\cos }^4}} \left( x \right){\sin ^2}\left( x \right)\sin \left( x \right)dx \cr & = \int_a^b {{{\cos }^4}} \left( x \right)\left( {1 - {{\cos }^2}\left( x \right)} \right)\sin \left( x \right)dx \cr & = \int_{\cos \left( a \right)}^{\cos \left( b \right)} - {u^4}\left( {1 - {u^2}} \right)du\,\,\,\,\,\,\left[ {{\text{Substituting: }}u = \cos \left( x \right)} \right] \cr & = - \int_{\cos \left( a \right)}^{\cos \left( b \right)} {{u^4}} du + \int_{\cos \left( a \right)}^{\cos \left( b \right)} {{u^6}} du \cr & = - \frac{{{{\cos }^5}\left( b \right)}}{5} + \frac{{{{\cos }^5}\left( a \right)}}{5} + \frac{{{{\cos }^7}\left( b \right)}}{7} - \frac{{{{\cos }^7}\left( a \right)}}{7} \cr} {/eq}

Solution (2):

Here, in this case, the given integral can be solved as:

{eq}\eqalign{ I& = \int_a^b {\frac{x}{{{x^2} + 4x + 1}}} dx \cr & = \int_a^b {\frac{x}{{{{\left( {x + 2} \right)}^2} - 3}}} dx \cr & = \int_{a + 2}^{b + 2} {\frac{{u - 2}}{{{u^2} - 3}}} du\,\,\,\,\,\,\,\left[ {{\text{Substitute: }}u = x + 2} \right] \cr & = \int_{a + 2}^{b + 2} {\frac{u}{{{u^2} - 3}}} du - \int_{a + 2}^{b + 2} {\frac{2}{{{u^2} - 3}}} du \cr & = \frac{1}{2}\int_{a + 2}^{b + 2} {\frac{{2u}}{{{u^2} - 3}}} du - 2\int_{a + 2}^{b + 2} {\frac{1}{{{u^2} - 3}}} du \cr & = \frac{1}{2}\left[ {\ln \left( {{u^2} - 3} \right)} \right]_{_{a + 2}}^{b + 2} - 2\frac{1}{{2\sqrt 3 }}\left[ {\ln \frac{{u - \sqrt 3 }}{{u + \sqrt 3 }}} \right]_{a + 2}^{b + 2} \cr & = \frac{1}{2}\left[ {\ln \left( {{{\left( {b + 2} \right)}^2} - 3} \right)} \right] - \frac{1}{2}\left[ {\ln \left( {{{\left( {a + 2} \right)}^2} - 3} \right)} \right] - \frac{1}{{\sqrt 3 }}\left[ {\ln \frac{{b + 2 - \sqrt 3 }}{{b + 2 + \sqrt 3 }}} \right] + \frac{1}{{\sqrt 3 }}\left[ {\ln \frac{{a + 2 - \sqrt 3 }}{{a + 2 + \sqrt 3 }}} \right] \cr} {/eq}

Solution (3):

Here, in this case, the given integral can be solved as:

{eq}\eqalign{ I& = \int_a^b c o{s^4}(x)si{n^4}(x)co{s^2}xdx \cr & = \frac{1}{{32}}\int_a^b {{{\left( {2cos(x)sin(x)} \right)}^4}} \left( {\cos 2x + 1} \right)dx \cr & = \frac{1}{{32}}\int_a^b {{{\left( {\sin 2x} \right)}^4}} \left( {\cos 2x + 1} \right)dx \cr & = \frac{1}{{32}}\left( {\int_a^b {{{\sin }^4}} \left( {2x} \right)\cos \left( {2x} \right) + {{\sin }^4}\left( {2x} \right)dx} \right) \cr & = \frac{1}{{32}}\left( {\frac{{{{\sin }^5}\left( {2b} \right) - {{\sin }^5}\left( {2a} \right)}}{{10}} + \frac{{ - 16\left( {\frac{{{{\sin }^3}\left( {2b} \right)}}{4}\cos \left( {2b} \right) - \frac{{{{\sin }^3}\left( {2a} \right)}}{4}\cos \left( {2a} \right)} \right) + 3\left( {4b - 4a - \sin \left( {4b} \right) + \sin \left( {4a} \right)} \right)}}{{32}}} \right) \cr} {/eq}


Learn more about this topic:

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Antiderivative: Rules, Formula & Examples

from Calculus: Help and Review

Chapter 8 / Lesson 12
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