Question

What is the value of \[\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } \]
A. \[\dfrac{1}{{12}}\]
B. \[\dfrac{3}{{12}}\]
C. \[\dfrac{5}{{12}}\]
D. None of these

Answer 1

Hint: We write the value of \[\sec \theta = \dfrac{1}{{\cos \theta }}\] and separate the terms in the numerator so we can apply the trigonometric identity \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]. Substituting the value of \[\cos \theta \] as a variable we change the integration in the form of the substituted variable. We solve the integration and put back the value of the variable as \[\cos \theta \] and then apply the limits.
* \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \]

Complete step-by-step answer:
We have the integration \[\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } \].
Substitute the value of \[\sec \theta = \dfrac{1}{{\cos \theta }}\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\sin }^3}\theta }}{{{{\cos }^7}\theta }}d\theta } \]
Now we write the numerator as \[{\sin ^3}\theta = {\sin ^2}\theta .\sin \theta \]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\sin }^2}\theta .\sin \theta }}{{{{\cos }^7}\theta }}d\theta } \]
Use the formula \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \] in the numerator, so we convert the whole term in the form of \[\cos \theta \].
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{(1 - {{\cos }^2}\theta ).\sin \theta }}{{{{\cos }^7}\theta }}d\theta } \]
Let us assume \[u = \cos \theta \]
Differentiating both sides of the equation
\[du = - \sin \theta d\theta \] { \[\dfrac{d}{{d\theta }}(\cos \theta ) = - \sin \theta \]}
Substitute the values in the integration without changing the limits.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} { - \dfrac{{(1 - {u^2})}}{{{u^7}}}du} \]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{u^2} - 1}}{{{u^7}}}du} \]
Solving RHS by multiplying numerator and denominator by \[{u^{ - 7}}\].
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{u^2} - 1}}{{{u^7}}} \times \dfrac{{{u^{ - 7}}}}{{{u^7}}}du} \]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{u^2}.{u^{ - 7}} - {u^{ - 7}}}}{{{u^7}.{u^{ - 7}}}}du} \]
Now we know that when the base is same powers get added
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{u^{2 - 7}} - {u^{ - 7}}}}{{{u^{7 - 7}}}}du} \]
Solving the values in power and using the fact that any number with power 0 is equal to one.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{u^{ - 5}} - {u^{ - 7}}}}{1}du} \]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \int\limits_0^{\dfrac{\pi }{4}} {{u^{ - 5}} - {u^{ - 7}}du} \]
Now using method of integration \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \]we solve RHS.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{u^{ - 5}} - {u^{ - 7}}du} = \left[ {\left( {\dfrac{{{u^{ - 5 + 1}}}}{{ - 5 + 1}}} \right) - \left( {\dfrac{{{u^{ - 7 + 1}}}}{{ - 7 + 1}}} \right)} \right]_0^{\dfrac{\pi }{4}}\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{u^{ - 5}} - {u^{ - 7}}du} = \left[ {\left( {\dfrac{{{u^{ - 4}}}}{{ - 4}}} \right) - \left( {\dfrac{{{u^{ - 6}}}}{{ - 6}}} \right)} \right]_0^{\dfrac{\pi }{4}}\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{u^{ - 5}} - {u^{ - 7}}du} = \left[ {\dfrac{1}{{6{u^6}}} - \dfrac{1}{{4{u^4}}}} \right]_0^{\dfrac{\pi }{4}}\]
Substitute the value of \[u = \cos \theta \]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\dfrac{1}{{6{{(\cos \theta )}^6}}} - \dfrac{1}{{4{{(\cos \theta )}^4}}}} \right]_0^{\dfrac{\pi }{4}}\]
Now applying the limits
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{1}{{6{{(\cos \dfrac{\pi }{4})}^6}}} - \dfrac{1}{{4{{(\cos \dfrac{\pi }{4})}^4}}}} \right) - \left( {\dfrac{1}{{6{{(\cos 0)}^6}}} - \dfrac{1}{{4{{(\cos 0)}^4}}}} \right)} \right]\]
We know \[\cos 0 = 1,\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{1}{{6{{(\dfrac{1}{{\sqrt 2 }})}^6}}} - \dfrac{1}{{4{{(\dfrac{1}{{\sqrt 2 }})}^4}}}} \right) - \left( {\dfrac{1}{{6{{(1)}^6}}} - \dfrac{1}{{4{{(1)}^4}}}} \right)} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{1}{{6(\dfrac{1}{8})}} - \dfrac{1}{{4(\dfrac{1}{4})}}} \right) - \left( {\dfrac{1}{{6(1)}} - \dfrac{1}{{4(1)}}} \right)} \right]\]
Solve the denominators.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{8}{6} - 1} \right) - \left( {\dfrac{1}{6} - \dfrac{1}{4}} \right)} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{4}{3} - 1} \right) - \left( {\dfrac{1}{6} - \dfrac{1}{4}} \right)} \right]\]
Take LCM on RHS
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{{4 - 3}}{3}} \right) - \left( {\dfrac{{4 - 6}}{{24}}} \right)} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{1}{3}} \right) - \left( {\dfrac{{ - 2}}{{24}}} \right)} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\left( {\dfrac{1}{3}} \right) + \left( {\dfrac{1}{{12}}} \right)} \right]\]
Take LCM again on RHS
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\dfrac{{4 + 1}}{{12}}} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \dfrac{5}{{12}}\]

So, option C is correct.

Note: Students make the mistake of not changing the sign when shifting values from one side to another. Also, the negative sign when we find the value of du should be multiplied when substituting the values of u and du. Always solve the calculations in the end step by step to avoid wrong answers.