Download App

INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int\text{x}^2\text{e}^{\text{x}^3}\cos\text{x}^3\text{dx}$
✓

Answer

Given integral is,
$\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\cos(\text{x}^3)\text{dx}$
Let $x^3 = t$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Integral becomes,
$\frac{1}{3}\int\text{e}^\text{t}\cos\text{t dt}$
$=\frac{1}{3}\text{I}\ \dots(1)$
Where, $\text{I}=\int\text{e}^\text{t}\cos\text{t dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{t dt}$
Considering cos t as first and $e^t$ as second function
$\text{I}=\cos\text{t e}^\text{t}-\int-\sin\text{t e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\text{e}^\text{t}\cos\text{t}+\int\sin\text{t }\text{e}^\text{t}\text{dt}$
Again considering sin t as first and $e^t$ as second function
$\text{I}=\text{e}^\text{t}\cos\text{t}+\sin\text{t }\text{e}^\text{t}-\int\cos\text{t e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\text{e}^\text{t}\cos\text{t}+\sin\text{t e}^\text{t}-1$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}}{2}(\sin\text{t}+\cos\text{t})$
$\therefore\ \int\text{x}^2\text{e}^{\text{x}^3}\cos(\text{x}^3)\text{dx}=\frac{1}{3}\Big[\frac{\text{e}^\text{t}}{2}(\sin\text{t}+\cos\text{t})\Big]+\text{C}$ [From (1)]
$=\frac{\text{e}^{\text{x}^3}}{6}(\sin\text{x}^3+\cos\text{x}^3)+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from INTEGRALS→INTEGRALS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Using vectors, find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
→
Prove that: $\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2},$where $\alpha=\text{ax}-\text{by}$ and $\beta=\text{ay}+\text{bx}.$
→
Minimise Z = -3x + 4 y
subject to $\text{x}+2\text{y}\leq8,\ 3\text{x}+2\text{y}\leq12,\ \text{x}\geq0,\ \text{y}\geq0.$
→
Evaluate the following integrals:
$\int(4\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
→
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,
$f(x) = x^3 - 2ax^3 + a^2x$
→
Maximize $Z = -x_1 + 2x_2$
Subject to
$-\text{x}_1+3\text{x}_2\leq10$
$\text{x}_1+\text{x}_2\leq6$
$\text{x}_1+\text{x}_2\leq2$
$\text{x}_1,\text{x}_2\geq0$
→
If $f(x) = x^2 - 2x$, find f(A), where $\text{A}=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
→
A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. Then, domain of R is:
  1. {2, 3, 5}
  2. {3, 5}
  3. {2, 3, 4}
  4. {2, 3, 4, 5}
→
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{\text{e}^\text{t}+\text{e}^{-\text{t}}}{2}\text{ and y}=\frac{\text{e}^\text{t}-\text{e}^\text{-t}}{2}$
→
Find a particular solution of the differential equation $\frac{ dy }{ dx }+2 y \cot x=4 x \operatorname{cosec} x$ $(x \neq 0)$, given that $y=0$, when $x=\frac{\pi}{2}$.
→