-4- -4loge sin x - cos x dx is

I = ∫^π/4_ π/4ln (√(2) sin(x + π/4))dx x + π/4 = θ⇒ dx = d θ I = ∫^π/2_0ln (√(2) sinθ)d θ = ∫^π/2_0ln√(2) d θ + ∫^π/2_0ln sinθ d θ = π/4ln 2 π/2ln 2 = π/4l ...

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

Byju's Answer

Standard XI

Mathematics

Integration Using Substitution

∫π/4- π/4loge...

Question

$\frac{π}{4} \int \frac{- π}{4} {log}_{e} (sin x + cos x) d x$ is

- \frac{π}{4} ln 2

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

\frac{π}{4} ln 2

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

\frac{π}{8} ln 2

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

- \frac{π}{8} ln 2

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

Open in App

Solution

The correct option is A $- \frac{π}{4} ln 2$
$I = \frac{π}{4} \int \frac{- π}{4} ln (\sqrt{2} sin (x + \frac{π}{4})) d x$
$x + \frac{π}{4} = θ \Rightarrow d x = d θ$
$I = \frac{π}{2} \int 0 ln (\sqrt{2} sin θ) d θ = \frac{π}{2} \int 0 ln \sqrt{2} d θ + \int_{0}^{\frac{π}{2}} ln sin θ d θ$
$= \frac{π}{4} ln 2 - \frac{π}{2} ln 2 = - \frac{π}{4} ln 2$

Suggest Corrections

Similar questions

Q. The value of

n π - \frac{π}{4} \int - \frac{π}{4} | sin x + cos x | d x

is:

\int_{π / 4}^{3 π / 4} |

cosx

|

=

Q. The value of

\int_{- \frac{π}{4}}^{n π - \frac{π}{4}} | sin x + cos x | d x

, is equal to

\int_{π / 4}^{3 π / 4} \frac{d x}{1 + cos x}

is equal to:

Q. The integral

\int_{\frac{π}{4}}^{\frac{3 π}{4}} \frac{d x}{1 + cos x}

Join BYJU'S Learning Program

π4∫−π4loge(sinx+cosx) dx is

The correct option is A −π4 ln2I=π4∫−π4ln(√2 sin(x+π4))dx x+π4=θ⇒dx=dθ I=π2∫0ln(√2 sinθ)dθ=π2∫0ln√2 dθ+∫π20ln sinθ dθ =π4ln2−π2ln2=−π4ln2

$\frac{π}{4} \int \frac{- π}{4} {log}_{e} (sin x + cos x) d x$ is