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Question: Which value of constant of integration makes the value of integral of sin 3x cos 5x equal to zero at...

Which value of constant of integration makes the value of integral of sin 3x cos 5x equal to zero at x = 0
A.316\dfrac{3}{{16}}
B.−316 - \dfrac{3}{{16}}
C.−56 - \dfrac{5}{6}
D.18\dfrac{1}{8}

Explanation

Solution

Hint: Integrations based questions are answered using directly the method of integration and the result will be equal to zero, using these instructions and method of integration i.e. ∫f(x)dx=f′(x)+C\int {f\left( x \right)dx} = f'\left( x \right) + Cwill help you to get a better approach towards the solution of the problem.

Complete step-by-step answer:
According to the given information we have function i.e. sin 3x cos 5x and the integration of given function is equal to 0 at x = 0
Before using the integration method on the given function let’s simplify the given function using the trigonometric formula sin⁡xcos⁡y=sin⁡(x+y)+sin⁡(x−y)2\sin x\cos y = \dfrac{{\sin \left( {x + y} \right) + \sin \left( {x - y} \right)}}{2}
By comparing the given function value of x and y are 3x and 5x
Substituting the values of given function in the above formula
sin⁡3xcos⁡5x=sin⁡(3x+5x)+sin⁡(3x−5x)2\sin 3x\cos 5x = \dfrac{{\sin \left( {3x + 5x} \right) + \sin \left( {3x - 5x} \right)}}{2}
⇒\Rightarrow sin⁡(8x)+sin⁡(−2x)2\dfrac{{\sin \left( {8x} \right) + \sin \left( { - 2x} \right)}}{2} Since sin (-x) = sin x
We get sin⁡8x−sin⁡2x2\dfrac{{\sin 8x - \sin 2x}}{2}equation 1
The general formula to find the integration of the function i.e. ∫f(x)dx=f′(x)+C\int {f\left( x \right)dx} = f'\left( x \right) + C
Now substituting the given function in equation 1 in the general formula of integration
I=∫(sin⁡8x−sin⁡2x2)dxI = \int {\left( {\dfrac{{\sin 8x - \sin 2x}}{2}} \right)dx} (equation 1)
Using the sum rule of integration in equation 1 which is ∫(f+g)dx=∫fdx+∫gdx\int {\left( {f + g} \right)dx} = \int {fdx} + \int {gdx} here f and g are 2 different functions
We get
I=12∫sin⁡8xdx−12∫sin⁡2xdxI = \dfrac{1}{2}\int {\sin 8xdx} - \dfrac{1}{2}\int {\sin 2xdx} (equation 2)
For integrating sin 8x
Let 8x = t
Differentiating both sides with respect to x
ddx8x=ddxt\dfrac{d}{{dx}}8x = \dfrac{d}{{dx}}t
⇒\Rightarrow 8ddxx=ddxt8\dfrac{d}{{dx}}x = \dfrac{d}{{dx}}t
Since we know that ddxx=1\dfrac{d}{{dx}}x = 1
We get 8=dtdx8 = \dfrac{{dt}}{{dx}}
⇒\Rightarrow dx=dt8dx = \dfrac{{dt}}{8}
By substituting the value of dx and 8x in equation 2
We get
I=12∫sin⁡tdt8−12∫sin⁡2xdxI = \dfrac{1}{2}\int {\sin t\dfrac{{dt}}{8}} - \dfrac{1}{2}\int {\sin 2xdx} (equation 3)
Now let cos t = u
Differentiating both side with respect to t
\Rightarrow $$$\dfrac{{d\cos t}}{{dt}} = \dfrac{{du}}{{dt}}$$ Since we know that $$\dfrac{{d\cos t}}{{dt}} = - \sin t$$ We get $$ - \sin t = \dfrac{{du}}{{dt}}$$ \Rightarrow dt = \dfrac{{du}}{{ - \sin t}}$$ Substituting the value of dt in equation 3 $$I = \dfrac{1}{{16}}\int {\sin t\dfrac{{du}}{{\left( { - \sin t} \right)}}} - \dfrac{1}{2}\int {\sin 2xdx} $$ $ \Rightarrow I = - \dfrac{1}{{16}}\int {du} - \dfrac{1}{2}\int {\sin 2xdx} (equation 4) For integrating sin 2x Let 2x = r Differentiating both side with respect to x $\dfrac{d}{{dx}}2x = \dfrac{d}{{dx}}r$ $ \Rightarrow $ $2\dfrac{d}{{dx}}x = \dfrac{d}{{dx}}r$ Since we know that $\dfrac{d}{{dx}}x = 1$ We get $2 = \dfrac{{dr}}{{dx}}$ $ \Rightarrow $ $dx = \dfrac{{dr}}{2}$ By substituting the value of dx in equation 4 We getI = - \dfrac{1}{{16}}\int {du} - \dfrac{1}{2}\int {\sin r\dfrac{{dr}}{2}} (equation5)Letcosr=vDifferentiatingbothsidewithrespecttor(equation 5) Let cos r = v Differentiating both side with respect to r \dfrac{{d\cos r}}{{dr}} = \dfrac{{dv}}{{dr}}Sinceweknowthat Since we know that\dfrac{{d\cos r}}{{dr}} = - \sin r $ \Rightarrow $$$ - \sin r = \dfrac{{dv}}{{dr}}
\Rightarrow $$$dr = \dfrac{{dv}}{{ - \sin r}}$$ Substituting the value of dr in equation 5 $$I = - \dfrac{1}{{16}}\int {du} - \dfrac{1}{4}\int {\sin r\dfrac{{dv}}{{\left( { - \sin r} \right)}}} $$ \Rightarrow I = - \dfrac{1}{{16}}\int {du} + \dfrac{1}{4}\int {dv} $$ Since according to the method of integration$\int {dx} = x$ $$I = - \dfrac{1}{{16}}u + \dfrac{1}{4}v + C$$ (equation 6) Substituting the values of u and v in equation 6 $$I = \dfrac{1}{{16}}\cos t - \dfrac{1}{4}\cos r + C$$ (equation 7) Now substituting the value of r in equation 7 $$I = \dfrac{1}{{16}}\cos 8x - \dfrac{1}{4}\cos 2x + C$$ Now we know that I = 0 at x = 0 Therefore $$I = \dfrac{1}{{16}}\cos 8\left( 0 \right) - \dfrac{1}{4}\cos 2\left( 0 \right) + C = 0$$ $ \Rightarrow I = \dfrac{1}{{16}}\cos 8\left( 0 \right) - \dfrac{1}{4}\cos 2\left( 0 \right) + C = 0 $ \Rightarrow $$$\dfrac{1}{{16}}\cos 0 - \dfrac{1}{4}\cos 0 + C = 0
Since cos 0 = 1
Therefore 116−14+C=0\dfrac{1}{{16}} - \dfrac{1}{4} + C = 0
\Rightarrow $$$\dfrac{-3}{{16}} + C = 0$$ \Rightarrow $C=316C = \dfrac{{ 3}}{{16}}
Hence option A is the correct option

Note: In the above solution the question was totally based on the concept of integration but do you know the basic mechanism behind the method of integration so let’s explain the concept of integration with help of an example suppose you have a banana and you have divided banana into so many small pieces so here the method of integration helps in adding those small pieces of banana to get the better result so mathematically you can say that the method of integration helps in adding lot smaller Δx\Delta xto find the whole measure of quantity for example integration is used to find the area, volume of irregular shapes or body.