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Question: Evaluate the following expression \(\int{{{\sin }^{5}}x.{{\cos }^{100}}xdx}\). a. \(\dfrac{{{\cos ...

Evaluate the following expression ∫sin⁡5x.cos⁡100xdx\int{{{\sin }^{5}}x.{{\cos }^{100}}xdx}.
a. cos⁡105x105+2cos⁡103x103−cos⁡101x101+c\dfrac{{{\cos }^{105}}x}{105}+2\dfrac{{{\cos }^{103}}x}{103}-\dfrac{{{\cos }^{101}}x}{101}+c
b. −cos⁡105x105+2cos⁡103x103−cos⁡101x101+c-\dfrac{{{\cos }^{105}}x}{105}+2\dfrac{{{\cos }^{103}}x}{103}-\dfrac{{{\cos }^{101}}x}{101}+c
c. −cos⁡105x105−2cos⁡103x103+cos⁡101x101+c-\dfrac{{{\cos }^{105}}x}{105}-2\dfrac{{{\cos }^{103}}x}{103}+\dfrac{{{\cos }^{101}}x}{101}+c
d. cos⁡105x105−2cos⁡103x103+cos⁡101x101+c\dfrac{{{\cos }^{105}}x}{105}-2\dfrac{{{\cos }^{103}}x}{103}+\dfrac{{{\cos }^{101}}x}{101}+c

Explanation

Solution

Hint: In order to solve this question, we should know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 and ddxcos⁡x=−sin⁡x\dfrac{d}{dx}\cos x=-\sin x. By using these concepts, we will try to form the whole integration in the form ∫(aun+bun+α+cun+β)du\int{\left( a{{u}^{n}}+b{{u}^{n+\alpha }}+c{{u}^{n+\beta }} \right)}du. And then we will simply apply ∫undu=un+1n+1+c\int{{{u}^{n}}du=\dfrac{{{u}^{n+1}}}{n+1}+c}.

Complete step-by-step answer:
In this question, we have been asked to calculate the value of ∫sin⁡5x.cos⁡100xdx\int{{{\sin }^{5}}x.{{\cos }^{100}}xdx}. To solve this question, we will first consider I=∫sin⁡5x.cos⁡100xdxI=\int{{{\sin }^{5}}x.{{\cos }^{100}}xdx}.
Now, we know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, so we can write sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x. Therefore, we get,
I=∫sin⁡x(1−cos⁡2x)2.cos⁡100xdxI=\int{\sin x{{\left( 1-{{\cos }^{2}}x \right)}^{2}}.{{\cos }^{100}}xdx}
Now, we will consider cos x = u. So, we can write – sin x dx = du. Therefore, we can write I as,
I=−∫(1−u2)2u100duI=-\int{{{\left( 1-{{u}^{2}} \right)}^{2}}{{u}^{100}}du}
Now, we know that (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab. So, for a = 1 and b = u2{{u}^{2}}, we get, (1−u2)2=1+u4−2u2{{\left( 1-{{u}^{2}} \right)}^{2}}=1+{{u}^{4}}-2{{u}^{2}}. Therefore, we get I as,
I=−∫(1+u4−2u2)u100duI=-\int{\left( 1+{{u}^{4}}-2{{u}^{2}} \right){{u}^{100}}du}
We can further write it as,

& I=-\int{\left( {{u}^{100}}+{{u}^{104}}-2{{u}^{102}} \right)du} \\\ & I=\int{\left( 2{{u}^{102}}-{{u}^{100}}-{{u}^{104}} \right)du} \\\ \end{aligned}$$ Now, we know that $\int{\left( a+b \right)dx=\int{adx+\int{bdx}}}$. So, we can write, $$I=\int{2{{u}^{102}}du-\int{{{u}^{100}}du}-\int{{{u}^{104}}du}}$$ And we can further write it as, $$I=2\int{{{u}^{102}}du-\int{{{u}^{100}}du}-\int{{{u}^{104}}du}}$$ Now, we know that $\int{{{u}^{n}}du=\dfrac{{{u}^{n+1}}}{n+1}+c}$. So, we can write I as, $I=2\times \dfrac{{{u}^{103}}}{103}-\dfrac{{{u}^{101}}}{101}-\dfrac{{{u}^{105}}}{105}+c$ And we can also write it as, $I=-\dfrac{{{u}^{105}}}{105}+2\times \dfrac{{{u}^{103}}}{103}-\dfrac{{{u}^{101}}}{101}+c$ Now, we will put the value of u, that is, u = cos x. So, we get I as, $I=-\dfrac{{{\cos }^{105}}x}{105}+2\dfrac{{{\cos }^{103}}x}{103}-\dfrac{{{\cos }^{101}}x}{101}+c$ Hence, we can say that, $\int{{{\sin }^{5}}x.{{\cos }^{100}}xdx}$ is equal to $-\dfrac{{{\cos }^{105}}x}{105}+2\dfrac{{{\cos }^{103}}x}{103}-\dfrac{{{\cos }^{101}}x}{101}+c$. Therefore, option (b) is the correct answer. Note: While solving this question, the possible mistake we can make is ignoring the negative sign of –sin x dx = du, which will give us the wrong answer. At times we do not get the hint of solving such questions, in that case we can try to get the question from differentiating the options given. By this method we will get the correct answer but it is a very time consuming process. Hence it is better to use the conventional method to solve such questions.