Write a value


Write a value of $\int e^{a x} \cos b x d x .5$


we know $\int f(x) g(x)=f(x) \int g(x)-\int f^{\prime}(x) \int g(x)$

Let $\int e^{a x} \cos b x d x=i$

Given that $\int e^{a x} \cos b x d x$

$i=\cos b x \int e^{a x}-\int-b \sin b x \int e^{a x}$

$i=\cos b x \frac{e^{a x}}{a}+\int b \sin b x \frac{e^{a x}}{a}$

$i=\cos b x \frac{e^{a x}}{a}+\frac{1}{a}\left[b \sin b x \frac{e^{a x}}{a}-\frac{b^{2}}{a} \int e^{a x} \cos b x d x\right]$

$\mathrm{I}=\cos b x \frac{e^{a x}}{a}+\frac{b}{a^{2}} \sin b x e^{a x}-\frac{b^{2}}{a^{2}} \mathrm{i}$

$\mathrm{i}\left(1+\frac{b^{2}}{a^{2}}\right)=\frac{a \cos b x e^{a x}+b \sin b x e^{a x}}{a^{2}}$

$\mathrm{i}=\frac{a \cos b x e^{a x}+b \sin b x e^{a x}}{a^{2}}\left(\frac{a^{2}}{a^{2}+b^{2}}\right)$

$\int e^{a x} \cos b x d x=\frac{e^{a x}(a \sin b x+b \cos b x)}{a^{2}+b^{2}}$

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