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Jaya Maheshwari 8 years, 2 months ago

The action or a process of differentiate or distinguish between two or more things.
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Ve
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Sia ? 4 years, 8 months ago

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Naveen Sharma 8 years, 2 months ago

The relation which equals the product of the sets on which it is defined is called trivial relation.

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Amar Kumar 8 years, 2 months ago

Linear differential equations are differential equations having solutions which can be added together in particular linear combinations to form further solutions. They equate 0 to a polynomial that is linear in the value and various derivatives of a variable; its linearity means that each term in the polynomial has degree either 0 or 1.

Linear differential equations are of the form Ly=f where the differential operator L is a linear operator, y is the unknown function, and the right hand side f is a given function (called the source term) with the same domain as y.

For functions dependent on t we may write the equation more expressly as

Ly(t)=f(t)

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Pranshu Tripathi 8 years, 2 months ago

On x-axis only there is vector i cap hence direction ratio is i
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Rahul Majhu 8 years, 2 months ago

Use formuka of tan of inverse
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Ankush Kumar 8 years, 2 months ago

O
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Naveen Sharma 8 years, 2 months ago

{tex}\int_\pi^0 {tanx \over sec x+ tan x}dx\\ = \int_\pi^0 {tanx \over sec x+ tan x}\times {sec x- tan x \over sec x- tan x} dx\\ = \int_\pi^0 {tanx ( sec x - tan x) \over sec ^2 x- tan^2 x} dx \\ = \int_\pi^0 {tanx ( sec x - tan x) } dx\\ = \int_\pi^0 {tanx.sec x - tan^2 x } dx\\ {/tex}

{tex}= \int_\pi^0 {tanx.sec x - ( sec^2 x-1) } dx\\ = \int_\pi^0 {tanx.sec x - sec^2 x+1 } dx\\ = \left | sec x - tanx + x \right |_\pi ^ 0 \\ = sec 0 - tan 0 + 1 - sec \pi + tan \pi - \pi \\ = 1 - 0 + 0 +1 + 0 - \pi \\ = 2 - \pi {/tex}

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