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प्रश्न
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उत्तर
\[\int\text{sin }\left( mx \right) \cdot \text{cos} \left( nx \right) dx\]
\[ = \frac{1}{2}\int2 \text{sin} \left( mx \right) \cdot \text{cos} \left( nx \right)dx\]
\[ = \frac{1}{2}\int\left[ \text{sin} \left( mx + nx \right) + \text{sin} \left( mx - nx \right) \right]dx \left[ \therefore \text{2 sin A }\cdot \text{cos B} = \text{sin} \left( A + B \right) + \text{sin} \left( A - B \right) \right]\]
\[ = \frac{1}{2}\left[ - \frac{\text{cos} \left( m + n \right)x}{m + n} - \frac{\text{cos} \left( m - n \right)x}{m - n} \right] + C\]
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संबंधित प्रश्न
Evaluate the following integrals:
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
