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Question: How do I evaluate \[\int {{{\cos }^5}(x){{\sin }^4}(x).dx} \] ?...

How do I evaluate ∫cos⁡5(x)sin⁡4(x).dx\int {{{\cos }^5}(x){{\sin }^4}(x).dx} ?

Explanation

Solution

Hint : The given function is indefinite since there is no limit given. The indefinite integral of a function is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integral. Here to solve this we use the Pythagoras identity of a trigonometry. That is sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 .

Complete step-by-step answer :
Given, ∫cos⁡5(x)sin⁡4(x).dx\int {{{\cos }^5}(x){{\sin }^4}(x).dx} .
Let choose,
I=∫cos⁡5(x)sin⁡4(x).dxI = \int {{{\cos }^5}(x){{\sin }^4}(x).dx} .
We can write cos⁡5(x)=cos⁡4(x).cos⁡(x){\cos ^5}(x) = {\cos ^4}(x).\cos (x) .
=∫sin⁡4(x)(cos⁡2(x))2cos⁡(x).dx= \int {{{\sin }^4}(x){{\left( {{{\cos }^2}(x)} \right)}^2}\cos (x).dx}
Now using the Pythagoras identity of a trigonometry we have cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x .
I=∫sin⁡4(x)(1−sin⁡2x)2cos⁡(x).dxI = \int {{{\sin }^4}(x){{\left( {1 - {{\sin }^2}x} \right)}^2}\cos (x).dx}
Now we use the substitute rule of trigonometry.
Let put sin⁡(x)=t\sin (x) = t . Differentiate it with respect to ‘x’.
⇒cos⁡(x)dx=dt\Rightarrow \cos (x)dx = dt
Substituting in the above integral.
Thus we have,
I=∫t4(1−t2)2dtI = \int {{t^4}{{\left( {1 - {t^2}} \right)}^2}dt}
We know (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab .
=∫t4(1+t4−2t2)dt= \int {{t^4}\left( {1 + {t^4} - 2{t^2}} \right)dt}
=∫(t4+t8−2t6)dt= \int {\left( {{t^4} + {t^8} - 2{t^6}} \right)dt}
We know ∫xn.dx=xn+1n+1+c\int {{x^n}.dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c , where ‘c’ is a integration constant.
Applying the integral we have,
=t4+14+1+t8+18+1−2t6+16+1+c= \dfrac{{{t^{4 + 1}}}}{{4 + 1}} + \dfrac{{{t^{8 + 1}}}}{{8 + 1}} - 2\dfrac{{{t^{6 + 1}}}}{{6 + 1}} + c
=t55+t99−2t77+c= \dfrac{{{t^5}}}{5} + \dfrac{{{t^9}}}{9} - 2\dfrac{{{t^7}}}{7} + c
Since we have sin⁡(x)=t\sin (x) = t , substituting this we have:
=(sin⁡(x))55+(sin⁡(x))99−2(sin⁡(x))77+c= \dfrac{{{{(\sin (x))}^5}}}{5} + \dfrac{{{{(\sin (x))}^9}}}{9} - 2\dfrac{{{{(\sin (x))}^7}}}{7} + c
Thus it can be written as
I=sin⁡5(x)5+sin⁡9(x)9−2sin⁡7(x)7+cI = \dfrac{{{{\sin }^5}(x)}}{5} + \dfrac{{{{\sin }^9}(x)}}{9} - 2\dfrac{{{{\sin }^7}(x)}}{7} + c , where ‘c’ is a integration constant. This is the required answer.
So, the correct answer is “ I=sin⁡5(x)5+sin⁡9(x)9−2sin⁡7(x)7+cI = \dfrac{{{{\sin }^5}(x)}}{5} + \dfrac{{{{\sin }^9}(x)}}{9} - 2\dfrac{{{{\sin }^7}(x)}}{7} + c ”.

Note : Don’t get confused with cos⁡n(x){\cos ^n}(x) and cos⁡(xn)\cos ({x^n}) . These both are different, we can write cos⁡n(x)=(cos⁡(x))n{\cos ^n}(x) = {(\cos (x))^n} . To simplify the given integral we use a substitution rule. Careful in the substitution rule, we need to put the result in terms of ‘x’ only. In an indefinite integral we have integration constant because we don’t have upper and lower limits in the integral. In a definite integral we do not have an integration constant because we have upper and lower limits in the integral. Careful in the calculation.