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Question: How do you integrate \(\int {{t^2} \cdot \cos \left( {1 - {t^3}} \right)} \,dt\)?...

How do you integrate ∫t2⋅cos⁡(1−t3) dt\int {{t^2} \cdot \cos \left( {1 - {t^3}} \right)} \,dt?

Explanation

Solution

Here in this question given an Indefinite integral, we have to find the integrated value of a given trigonometric function. This can be solved by the substitution method and later integrated by using the standard trigonometric formula of integration. And by further simplification we get the required solution.

Complete step by step answer:
Integration is the inverse process of differentiation. An integral which does not have any upper and lower limit is known as an indefinite integral.Consider the given function.
⇒   ∫t2⋅cos⁡(1−t3) dt\Rightarrow \,\,\,\int {{t^2} \cdot \cos \left( {1 - {t^3}} \right)} \,dt
On rearranging, we can also written the above equation as
⇒   ∫cos⁡(1−t3)⋅t2 dt\Rightarrow \,\,\,\int {\cos \left( {1 - {t^3}} \right)} \cdot {t^2}\,dt --------- (1)
Apply the method of substitution
put 1−t3=u1 - {t^3} = u, then
On differentiating u with respect to t, we get
−3t2=dudt- 3{t^2} = \dfrac{{du}}{{dt}}
Or it can be written as
−3t2dt=du- 3{t^2}dt = du or t2dt=−13du{t^2}dt = - \dfrac{1}{3}du.

Then, Integral (1) becomes
⇒   ∫cos⁡u⋅(−13)du\Rightarrow \,\,\,\int {\cos u} \cdot \left( { - \dfrac{1}{3}} \right)du
Take constant outside the integral sign
⇒ − 13∫cos⁡u⋅du\Rightarrow \, - \,\dfrac{1}{3}\int {\cos u} \cdot du------- (2)
By the standard integration formula ∫cos⁡x dx=−sin⁡x+C\int {\cos x\,dx} = - \sin x + C, then
Integrate Equation (2) with respect to u.
⇒ −13⋅(sin⁡u)+C\Rightarrow \, - \dfrac{1}{3} \cdot \left( {\sin u} \right) + C
Substitute the u value as u=1−t3u = 1 - {t^3}, then
⇒ −13⋅sin⁡(1−t3)+C\Rightarrow \, - \dfrac{1}{3} \cdot \sin \left( {1 - {t^3}} \right) + C.
Where, C is an integrating constant.

Hence,the integration of ∫t2⋅cos⁡(1−t3) dt\int {{t^2} \cdot \cos \left( {1 - {t^3}} \right)} \,dt is equal to
−13⋅sin⁡(1−t3)+C- \dfrac{1}{3} \cdot \sin \left( {1 - {t^3}} \right) + C.**

Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must be known.