The value of -4-4lnsin x-cos xdx is-

Let I=∫_ π/4^π/4ln(sin x+cos x)dx ⇒ I =∫_ π/4^π/4ln [√(2)sin ( x+π4 ) ] dx Putting x+π4=θ⇒ dx=dθ ⇒ I =∫_0^π/2ln(√(2)sinθ) dθ =12∫_0^π/2ln 2 dθ+∫_0^π/2lns ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The value of ...

Question

The value of $\frac{π}{4} \int - \frac{π}{4} ln (sin x + cos x) d x$ is:

\frac{π}{2} ln \frac{1}{2}

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\frac{π}{4} ln 2

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\frac{π^{2}}{2} ln \frac{1}{2}

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\frac{π}{4} ln \frac{1}{2}

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Solution

The correct option is D $\frac{π}{4} ln \frac{1}{2}$
Let $I = \frac{π}{4} \int - \frac{π}{4} ln (sin x + cos x) d x$
$\Rightarrow I = \frac{π}{4} \int - \frac{π}{4} ln [\sqrt{2} sin (x + \frac{π}{4})] d x$
$Putting x + \frac{π}{4} = θ \Rightarrow d x = d θ$
$\Rightarrow I = \frac{π}{2} \int 0 ln (\sqrt{2} sin θ) d θ = \frac{1}{2} \frac{π}{2} \int 0 ln 2 d θ + \frac{π}{2} \int 0 ln sin θ d θ = \frac{π}{4} ln 2 - \frac{π}{2} ln 2 = - \frac{π}{4} ln 2 ∴ I = \frac{π}{4} ln \frac{1}{2}$

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Similar questions

I_{1} = \int_{0}^{\frac{π}{2}} l n (s i n x) d x, I_{2} = \int_{- \frac{π}{4}}^{\frac{π}{4}} ln (sin x + cos x) d x

.Then

Q. If

I = \int_{0}^{π / 2} ℓ n (sin x) d x

then

\int_{- π / 4}^{π / 4} ℓ n (sin x + cos x) d x =

$\int_{\frac{π}{4}}^{\frac{π}{4}} l o g (s i n x + c o s x) d x$

Q. The value of integral

\int_{π_{/ 4}}^{π / 2} cos x d x

is?

Q. The value of

\int_{π / 4}^{π / 2} e^{x} (log sin x + cot x) d x

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Second Fundamental Theorem of Calculus

Standard XII Mathematics

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The value of π4∫−π4ln(sinx+cosx)dx is:

The correct option is D π4ln12Let I=π4∫−π4ln(sinx+cosx)dx ⇒I=π4∫−π4ln[√2sin(x+π4)] dx Putting x+π4=θ⇒dx=dθ ⇒I=π2∫0ln(√2sinθ) dθ =12π2∫0ln2 dθ+π2∫0lnsinθ dθ =π4ln2−π2ln2=−π4ln2∴I=π4ln12

The value of $\frac{π}{4} \int - \frac{π}{4} ln (sin x + cos x) d x$ is: