Let J=∫−5−4 3−x2tan⁡3−x2dx andK=∫−2−1 6−6x+x2tan⁡6x−x2−6dx. Then (J+K) equals __.

# Let $J={\int }_{-5}^{-4} \left(3-{x}^{2}\right)\mathrm{tan}\left(3-{x}^{2}\right)dx$ and$K={\int }_{-2}^{-1} \left(6-6x+{x}^{2}\right)\mathrm{tan}\left(6x-{x}^{2}-6\right)dx$. Then $\left(J+K\right)$ equals __.

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### Solution:

We have $J={\int }_{-5}^{-4} \left(3-{x}^{2}\right)\mathrm{tan}\left(3-{x}^{2}\right)dx$

Put $\left(x+5\right)=t$. Then

$\begin{array}{l}J={\int }_{0}^{1} \left(3-\left(t-5{\right)}^{2}\right)\mathrm{tan}\left(3-\left(t-5{\right)}^{2}\right)dt\\ ={\int }_{0}^{1} \left(-22+10t-{t}^{2}\right)\mathrm{tan}\left(-22+10t-{t}^{2}\right)dt\end{array}$

Now $K={\int }_{-2}^{-1} \left(6-6x+{x}^{2}\right)\mathrm{tan}\left(6x-{x}^{2}-6\right)dx$.

Put $\left(x+2\right)=z$. Then

$\begin{array}{l}K={\int }_{0}^{1} \left(6-6\left(z-2\right)+\left(z-2{\right)}^{2}\right)\mathrm{tan}\left(6\left(z-2\right)-\left(z-2{\right)}^{2}-6\right)dz\\ ={\int }_{0}^{1} \left(22-10z+{z}^{2}\right)\mathrm{tan}\left(-22+10z-{z}^{2}\right)dz\end{array}$

Hence, $\left(J+K\right)=0$

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