-12-16n2-4sin-x-x d x-

The correct option is B √(2)Put √(x)=t ⇒1/2 √(x)dx=dt x=π^2/16⇒ t=π/4,x=π^2/4⇒ t=π/2 ∫_π^2/16^π^2/4 sin √(x)/√(x)dx=∫_π/4^π/2 [sin t. 2 dt= 2 cost]∫_π/4^π/2= 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Property 1

∫12/16n2/4sin...

Question

$\int_{\frac{π^{2}}{16}}^{\frac{π^{2}}{4}} \frac{s i n \sqrt{x}}{\sqrt{x}} d x =$

\sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\sqrt{2}$
Put $\sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t$
$x = \frac{π^{2}}{16} \Rightarrow t = \frac{π}{4}, x = \frac{π^{2}}{4} \Rightarrow t = \frac{π}{2}$
$\int_{\frac{π^{2}}{16}}^{\frac{π^{2}}{4}} \frac{s i n \sqrt{x}}{\sqrt{x}} d x = \int_{\frac{π}{4}}^{\frac{π}{2}} [s i n t . 2 d t = - 2 c o s t] \int_{\frac{π}{4}}^{\frac{π}{2}} = - 2 [0 - \frac{1}{\sqrt{2}}] = \sqrt{2}$

Suggest Corrections

Similar questions

\int_{π^{2} / 16}^{π^{2} / 4} \frac{sin \sqrt{x}}{\sqrt{x}} d x =

Q. The value of the integral

\int_{0}^{\frac{π^{2}}{4}} sin \sqrt{x} d x

\int_{1}^{2} \sqrt{(x - 1) (2 - x)} d x =

\frac{9 π}{8} - \frac{9}{4} {sin}^{- 1} \frac{1}{3} = \frac{9}{4} {sin}^{- 1} \frac{2 \sqrt{2}}{3}

Q. Prove:

\frac{9 π}{8} - \frac{9}{4} {sin}^{- 1} \frac{1}{3} = \frac{9}{4} {sin}^{- 1} \frac{2 \sqrt{2}}{3}

Join BYJU'S Learning Program

∫π24π216 sin√x√xdx=

The correct option is A √2Put √x=t⇒12√xdx=dt x=π216⇒t=π4,x=π24⇒t=π2 ∫π24π216 sin√x√xdx=∫π2π4[sin t. 2 dt=−2 cost]∫π2π4=−2[0−1√2]=√2

$\int_{\frac{π^{2}}{16}}^{\frac{π^{2}}{4}} \frac{s i n \sqrt{x}}{\sqrt{x}} d x =$