What is h(t-a)e^{-bt}cos^2(t-a) ?

The solution to h(t-a)e^{-bt}cos^2(t-a) is (hcos^2(t-a)(t-a))/(e^{bt)}

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simplify h(t-a)e^{-bt}cos^2(t-a)

simplify

h (t - a) e^{- b t} cos^{2} (t - a)

\frac{h cos ^{2} ( t - a ) ( t - a )}{e ^{b t}}

Solution steps

h (t - a) e^{- b t} cos^{2} (t - a)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{b t}} h cos^{2} (t - a) (t - a)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot h ( t - a ) cos ^{2} ( t - a )}{e ^{b t}}

Multiply:

1 \cdot h = h

= \frac{h cos ^{2} ( t - a ) ( t - a )}{e ^{b t}}

e^{- (l n (x + 1) + x)}

x^{2} (3 x^{2} - 3)^{2}

e^{- \frac{r t}{l}}

t \cdot e^{a} t \cdot cos (b t)

What is h(t-a)e^{-bt}cos^2(t-a) ?
The solution to h(t-a)e^{-bt}cos^2(t-a) is (hcos^2(t-a)(t-a))/(e^{bt)}