Advertisements
Advertisements
प्रश्न
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
योग
Advertisements
उत्तर
Let f(x) = `x^3 cos^3x`
f(– x) = `(- x)^3 cos^3 (- x)`
= `- x^3 cos^3 x`
= `- "f"(x)`
Here f(– x) = – f(x)
∴ f(x) is an odd function
∴ `int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x` = 0
shaalaa.com
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]
\[\int\limits_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} dx\]
\[\int_0^\pi e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx\]
\[\int\limits_0^1 x e^{x^2} dx\]
\[\int\limits_0^{\pi/2} x^2 \sin\ x\ dx\]
\[\int_0^2 2x\left[ x \right]dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\] is equal to
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
