Advertisements
Advertisements
प्रश्न
Evaluate the following integrals:
उत्तर
`"sec x"/("sec"(2"x")) = ("cos"(2"x"))/"cos x"`
`= (2 "cos"^2"x" - 1)/"cos x"`
`= ("2 cos"^"x")/"cos x" - 1/"cos x"`
= 2 cos x - sec x
`int ("sec"("x"))/("sec"(2"x")) "dx" = int[2 "cos x" - "sec x"] "dx"`
`= 2 int "cos x" "dx" - int "sec"("x") "dx"`
= 2 sin(x) - ln |sec (x) + tan (x)| + C
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^3dx/(9+x^2)`
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
