Advertisements
Advertisements
प्रश्न
Find: ∫ sin x · log cos x dx
Advertisements
उत्तर
`∫ sin "x" ·log cos"x" "dx"`
Substitute cos x = t
sinx dx = dt
`∫ - "log t dt"`
= - (t log t - t ) + C
= - t log t + t + C
= - cos x log (cos x )+ cos x + C
APPEARS IN
संबंधित प्रश्न
If `sin (sin^(−1)(1/5)+cos^(−1) x)=1`, then find the value of x.
Prove that:
`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`
Prove `tan^(-1) 2/11 + tan^(-1) 7/24 = tan^(-1) 1/2`
Write the function in the simplest form: `tan^(-1) ((cos x - sin x)/(cos x + sin x)) `,` 0 < x < pi`
Write the following function in the simplest form:
`tan^(-1) x/(sqrt(a^2 - x^2))`, |x| < a
Find the value of the following:
`tan 1/2 [sin^(-1) (2x)/(1+ x^2) + cos^(-1) (1-y^2)/(1+y^2)]`, |x| < 1, y > 0 and xy < 1
Find the value of the given expression.
`tan^(-1) (tan (3pi)/4)`
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
If tan-1 x - cot-1 x = tan-1 `(1/sqrt(3)),`x> 0 then find the value of x and hence find the value of sec-1 `(2/x)`.
Solve: tan-1 4 x + tan-1 6x `= π/(4)`.
Find the value of the expression in terms of x, with the help of a reference triangle
cos (tan–1 (3x – 1))
Prove that `tan^-1 2/11 + tan^-1 7/24 = tan^-1 1/2`
Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`
Choose the correct alternative:
If `sin^-1x + cot^-1 (1/2) = pi/2`, then x is equal to
Prove that `2sin^-1 3/5 - tan^-1 17/31 = pi/4`
Prove that cot–17 + cot–18 + cot–118 = cot–13
Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`
If `tan^-1x = pi/10` for some x ∈ R, then the value of cot–1x is ______.
If `sin^-1 ((2"a")/(1 + "a"^2)) + cos^-1 ((1 - "a"^2)/(1 + "a"^2)) = tan^-1 ((2x)/(1 - x^2))`. where a, x ∈ ] 0, 1, then the value of x is ______.
The value of the expression `tan (1/2 cos^-1 2/sqrt(5))` is ______.
If tan-1 2x + tan-1 3x = `pi/4,` then x is ____________.
`"cos"^-1["cos"(2"cot"^-1(sqrt2 - 1))]` = ____________.
`"sin"^-1 ((-1)/2)`
If `"sin"^-1 (1 - "x") - 2 "sin"^-1 ("x") = pi/2,` then x is equal to ____________.
The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:
𝐴' Is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠CAB and ∠CA'B is ______.
`tan^-1 1/2 + tan^-1 2/11` is equal to
The Simplest form of `cot^-1 (1/sqrt(x^2 - 1))`, |x| > 1 is
The value of cosec `[sin^-1((-1)/2)] - sec[cos^-1((-1)/2)]` is equal to ______.
