Integrate.

Calculus Level 3

To 3 decimal places, evaluate

0 π / 2 ln ( sin ( x ) ) d x . \int _{ 0 }^{ { \pi }/{ 2 } }{ \ln { \left( \sin { \left( x \right) } \right) } }\, dx.


The answer is -1.089.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Avineil Jain
Jun 13, 2014

I = 0 π / 2 l n ( s i n x ) d x I = \displaystyle\int_{0}^{π/2}ln(sin~x)dx

I = 0 π / 2 l n ( c o s x ) d x I = \displaystyle\int_{0}^{π/2}ln(cos~x)dx

Adding,

2 I = 0 π / 2 l n ( s i n 2 x 2 ) d x 2I = \displaystyle\int_{0}^{π/2}ln(\dfrac{sin~2x}{2})dx

2 I = 0 π / 2 l n ( s i n 2 x ) l n 2 d x 2I = \displaystyle\int_{0}^{π/2}ln(sin~2x) - ln2~dx

2 I = 0 π / 2 l n ( s i n 2 x ) d x π 2 l n 2 2I = \displaystyle\int_{0}^{π/2}ln(sin~2x)dx - \dfrac{\pi}{2}ln2

Let 2 x = t 2x=t . Then, 2 d x = d t 2dx = dt

2 I = 1 2 0 π l n ( s i n t ) d t π 2 l n 2 2I = \dfrac{1}{2}\displaystyle\int_{0}^{π}ln(sin~t)dt - \dfrac{\pi}{2}ln2

2 I = 0 π / 2 l n ( s i n t ) d t π 2 l n 2 2I = \displaystyle\int_{0}^{π/2}ln(sin~t)dt - \dfrac{\pi}{2}ln2

2 I = I π 2 l n 2 2I = I - \dfrac{\pi}{2}ln2

I = π 2 l n 2 I = -\dfrac{\pi}{2}ln2

I = 1.089 I = -1.089

how is ln sinx and ln cos x equal that u'hv added these two?

AnWesa Royce - 6 years, 12 months ago

Log in to reply

the area under the curves from 0 to pi/2 is symmetric

RYan Joseph - 6 years, 11 months ago

got it from suddep's explanation :)

AnWesa Royce - 6 years, 12 months ago

I also used the exact approach but got it wrong at the end

Ronak Agarwal - 6 years, 12 months ago
Sudeep Salgia
Jun 13, 2014

Using properties of definite integrals it can easily be evaluated. I will use the following properties:

  • 0 a f ( x ) d x = 0 a f ( a x ) d x \displaystyle \int \limits_0^a f(x) \text{d}x = \int \limits_0^a f(a-x) \text{d}x

  • 0 2 a f ( x ) d x = 2 × 0 a f ( x ) d x , if f ( x ) = f ( 2 a x ) \displaystyle \int \limits_0^{2a} f(x) \text{d}x = 2 \times \int \limits_0^a f(x) \text{d}x \text{ , if } f(x) = f(2a - x)

  • a b f ( x ) d x = a b f ( t ) d t \displaystyle \int \limits_a^b f(x) \text{d}x = \int \limits_a^b f(t) \text{d}t

Let I = 0 π 2 l n ( sin x ) d x I = 0 π 2 l n ( cos x ) d x \displaystyle I = \int \limits_0^{\frac{\pi }{2}} ln(\sin x) \text{d}x \text{ } \Rightarrow I = \int \limits_0^{\frac{\pi }{2}} ln(\cos x) \text{d}x

2 I = 0 π 2 ( l n ( sin x ) + l n ( cos x ) ) d x 2 I = 0 π 2 l n ( sin x cos x ) d x \displaystyle \Rightarrow 2I = \int \limits_0^{\frac{\pi }{2}} (ln(\sin x) + ln(\cos x)) \text{d}x \text{ } \Rightarrow 2I = \int \limits_0^{\frac{\pi }{2}} ln(\sin x \text{ } \cos x) \text{d}x

2 I = 0 π 2 l n ( sin 2 x ) d x 0 π 2 l n ( 2 ) d x \displaystyle \Rightarrow 2I = \int \limits_0^{\frac{\pi }{2}} ln(\sin 2x) \text{d}x - \int \limits_0^{\frac{\pi }{2}} ln(2) \text{d}x

Put 2 x = t 2x = t to get,
2 I = 1 2 0 π l n ( sin t ) d t π 2 l n 2 \displaystyle 2I =\frac{1}{2} \int \limits_0^{\pi } ln(\sin t) \text{d}t - \frac{\pi }{2} ln\text{ } 2

2 I = 2 2 0 π 2 l n ( sin t ) d t π 2 l n 2 \displaystyle \Rightarrow 2I = \frac{2}{2} \int \limits_0^{\frac{\pi }{2}} ln(\sin t) \text{d}t - \frac{\pi }{2} ln\text{ } 2

2 I = I π 2 l n 2 \displaystyle \Rightarrow 2I = I - \frac{\pi }{2} ln\text{ }2

Therefore,
0 π 2 l n ( sin x ) d x = π 2 l n 2 = 1.08879 \displaystyle \boxed{ \int \limits_0^{\frac{\pi }{2}} ln(\sin x) \text{d}x = - \frac{\pi }{2} ln\text{ }2 = - 1.08879}

You wouldn't believe but we wrote the solutions at nearly identical times!

Avineil Jain - 6 years, 12 months ago

Log in to reply

Not only that but our solutions are same word to word too. :D

Sudeep Salgia - 6 years, 12 months ago

Log in to reply

That is true! :)

Avineil Jain - 6 years, 12 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...