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Question: What is the value of \[\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta ...

What is the value of ∫0π4sec⁡7θsin⁡3θdθ\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta }
A. 112\dfrac{1}{{12}}
B. 312\dfrac{3}{{12}}
C. 512\dfrac{5}{{12}}
D. None of these

Explanation

Solution

We write the value of sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }} and separate the terms in the numerator so we can apply the trigonometric identity sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta . Substituting the value of cos⁡θ\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⁡θ\cos \theta and then apply the limits.

  • ∫xndx=xn+1n+1\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}

Complete step-by-step answer:
We have the integration ∫0π4sec⁡7θsin⁡3θdθ\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } .
Substitute the value of sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }}
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4sin⁡3θcos⁡7θdθ\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⁡3θ=sin⁡2θ.sin⁡θ{\sin ^3}\theta = {\sin ^2}\theta .\sin \theta
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4sin⁡2θ.sin⁡θcos⁡7θdθ\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⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta in the numerator, so we convert the whole term in the form of cos⁡θ\cos \theta .
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4(1−cos⁡2θ).sin⁡θcos⁡7θdθ\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⁡θu = \cos \theta
Differentiating both sides of the equation
du=−sin⁡θdθdu = - \sin \theta d\theta { ddθ(cos⁡θ)=−sin⁡θ\dfrac{d}{{d\theta }}(\cos \theta ) = - \sin \theta }
Substitute the values in the integration without changing the limits.
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4−(1−u2)u7du\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}
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4u2−1u7du\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{u^{ - 7}}.
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4u2−1u7×u−7u7du\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}
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4u2.u−7−u−7u7.u−7du\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
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4u2−7−u−7u7−7du\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.
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4u−5−u−71du\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}
⇒∫0π4sec⁡7θsin⁡3θdθ=∫0π4u−5−u−7du\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 ∫xndx=xn+1n+1\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} we solve RHS.
⇒∫0π4u−5−u−7du=[(u−5+1−5+1)−(u−7+1−7+1)]0π4\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}}
⇒∫0π4u−5−u−7du=[(u−4−4)−(u−6−6)]0π4\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}}
⇒∫0π4u−5−u−7du=[16u6−14u4]0π4\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⁡θu = \cos \theta
⇒∫0π4sec⁡7θsin⁡3θdθ=[16(cos⁡θ)6−14(cos⁡θ)4]0π4\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
⇒∫0π4sec⁡7θsin⁡3θdθ=[(16(cos⁡π4)6−14(cos⁡π4)4)−(16(cos⁡0)6−14(cos⁡0)4)]\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⁡π4=12\cos 0 = 1,\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}
⇒∫0π4sec⁡7θsin⁡3θdθ=[(16(12)6−14(12)4)−(16(1)6−14(1)4)]\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]
⇒∫0π4sec⁡7θsin⁡3θdθ=[(16(18)−14(14))−(16(1)−14(1))]\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.
⇒∫0π4sec⁡7θsin⁡3θdθ=[(86−1)−(16−14)]\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]
⇒∫0π4sec⁡7θsin⁡3θdθ=[(43−1)−(16−14)]\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
⇒∫0π4sec⁡7θsin⁡3θdθ=[(4−33)−(4−624)]\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]
⇒∫0π4sec⁡7θsin⁡3θdθ=[(13)−(−224)]\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]
⇒∫0π4sec⁡7θsin⁡3θdθ=[(13)+(112)]\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
⇒∫0π4sec⁡7θsin⁡3θdθ=[4+112]\Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^7}\theta {{\sin }^3}\theta d\theta } = \left[ {\dfrac{{4 + 1}}{{12}}} \right]
⇒∫0π4sec⁡7θsin⁡3θdθ=512\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.