-0-4dx-k- Then the value of k-

The correct option is A π 4 ∫_0^π/41dx =[x]_0^π/4= (π/4 0)=π4 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Fractional Exponent

∫0π/4dx=k. Th...

Question

$\int_{0}^{π / 4} d x = k$ . Then the value of k=

\frac{π}{4}

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

\frac{π}{2}

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

\frac{3 π}{4}

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

none of these

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

Open in App

Solution

Suggest Corrections

Similar questions

\int_{0}^{π / 4} \sqrt{tan θ} d θ = \frac{1}{\sqrt{(2)}} l o g (\sqrt{2} - 1) + \frac{π}{2 k \sqrt{(2)}}

k

\int_{0}^{π / 2} \frac{cos x d x}{1 - {sin}^{2} x + {sin}^{4} x} = \frac{π}{k} + \frac{1}{4 \sqrt{(3)}} l o g (\frac{(2 + \sqrt{3})}{(2 - \sqrt{3})})

k

3 π / 4 \int - π / 4 \frac{e^{π / 4} d x}{(e^{x} + e^{π / 4}) (sin x + cos x)} = k π / 2 \int - π / 2 sec x d x

k

\int_{- π / 4}^{3 π / 4} \frac{e^{π / 4} d x}{(e^{x} + e^{π / 4}) (sin x + cos x)} = k \int_{- \frac{π}{2}}^{\frac{π}{2}} sec x d x

\int_{0}^{π / 2} log sin x d x = \frac{π}{k} log \frac{1}{2} .

k

Join BYJU'S Learning Program