What is the integral of 10^{tan(x)+1000}sec^2(x) ?

The integral of 10^{tan(x)+1000}sec^2(x) is (10^{tan(x)+1000})/(ln(10))+C

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integral of 10^{tan(x)+1000}sec^2(x)

\int 1 0^{t a n (x) + 1000} sec^{2} (x) d x

\frac{1 0 ^{t a n (x) + 1000}}{ln ( 10 )} + C

Solution steps

\int 1 0^{t a n (x) + 1000} sec^{2} (x) d x

1 0^{t a n (x) + 1000} sec^{2} (x) = 1 0^{1000} \cdot 1 0^{t a n (x)} sec^{2} (x)

= \int 1 0^{1000} \cdot 1 0^{t a n (x)} sec^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1 0^{1000} \cdot \int 1 0^{t a n (x)} sec^{2} (x) d x

Apply u-substitution

: \int 1 0^{u} d u

= 1 0^{1000} \cdot \int 1 0^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 1 0^{1000} \cdot \frac{1 0 ^{u}}{ln ( 10 )}

Substitute back

u = tan (x)

= 1 0^{1000} \cdot \frac{1 0 ^{t a n (x)}}{ln ( 10 )}

Simplify

1 0^{1000} \cdot \frac{1 0 ^{t a n (x)}}{ln ( 10 )} : \frac{1 0 ^{t a n (x) + 1000}}{ln ( 10 )}

= \frac{1 0 ^{t a n (x) + 1000}}{ln ( 10 )}

Add a constant to the solution

= \frac{1 0 ^{t a n (x) + 1000}}{ln ( 10 )} + C

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What is the integral of 10^{tan(x)+1000}sec^2(x) ?
The integral of 10^{tan(x)+1000}sec^2(x) is (10^{tan(x)+1000})/(ln(10))+C