Evaluate the following integrals: ^ _0x ^4x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}$

✓

Answer

Let $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^4(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^4\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin\text{x}\cos^4\text{x dx}$
$=\pi\int\limits^{\pi}_0\sin\text{x}\cos^4\text{x dx}$
Let $\cos\text{x}=\text{t},$ Then $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1,\text{x}=\pi,\text{t}=-1$
Therefore, $2\text{I}=-\pi\int\limits^{-1}_1\text{t}^4\text{ dt}$
$=\pi\int\limits^{1}_{-1}\text{t}^4\text{ dt}$
$=\pi\Big[\frac{\text{t}^5}{5}\Big]^{1}_{-1}$
$=\frac{\pi}{5}+\frac{\pi}{5}$
$=\frac{2\pi}{5}$
Hence, $\text{I}=\frac{\pi}{5}$

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 "4 Marks" from Definite Integrals→Definite Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Consider the binary operations *: R × R → R and o: R × R → R defined as $\text{a}∗\text{b} = |\text{a } – \text{ b}|\text{ and}\text{ a} o \text{b} = \text{a},\forall\text{a},\text{b}\in\text{R}.$ Show that * is commutative but not associative, o is associative but not commutative. Further, show that $\forall\text{a},\text{b},\text{c}\in\text{R},\text{a} *(\text{b}o\text{c}) = (\text{a} *\text{b})o(\text{a}*\text{b}).$ [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

Find the value of a and b so that the function f(x) defind by $\text{f(x)}=\begin{cases}\text{x}+\text{a}\sqrt{2}\sin\text{x},&\text{if }0\leq\text{x}<\frac{\pi}{4}\\2\text{x}\cot\text{ x}+\text{b},&\text{if }\frac{\pi}{4}\leq\text{x}<\frac{\pi}{2}\\\text{a}\cos2\text{x}-\text{b}\sin\text{x},&\text{if }\frac{\pi}{2}\leq\text{x}\leq\pi\end{cases}$ becomes continuous on $[0,\pi]$

If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$

A square piece of tin of side 18cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume.

A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10m/ s, how fast is the string being let out; when the kite is 250m away from the boy who is flying the kite? The height of boy is 1.5m.

Explain if Rolle's theorem is applicable to any one of the following functions.

$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$

Can you say something about the converse of Rolle's Theorem from these functions?

Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$

Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.

Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$

If $\text{y}=\tan^{-1}\text{x},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y alone.